I'm going to try to ask an actually interesting and well-researched question here, so this will be a long post.

Also, I recommend you read this MO post for context, since it contains lots of background and context for this question.

Since this is a long post, I assume most people won't want to read the whole thing-- if your willing to grant that the method works fairly generally and is useful, I'd recommend

Background

An enormously useful trick in analyzing an infinite series of the form \( \sum_{n=1}^\infty f(n) \) is to expand \( f(n) \) as its own infinite series, and then swap the order of summation, and sum the inner series by analytical continuation. If you read the linked MO post, then you saw my example where the naive approach of simply replacing the inner series by analytical continuation fails. My solution to this problem was to pick up the residues of the analytical continuation of the inner series. In doing some recent calculations, I've come to realize that my solution doesn't quite work-- but I think its definitely on the right track. Thus, to start, let me illustrate some cases where my theory produces some useful results.

Case 1: Euler-Maclaurin Formula

Let us assume that \( f(n) \) can be represented by a power series at 0. Then \(f(n) = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} n^k \). Now, lets try to analyze the series \( \sum_{n=1}^\infty f(n)\). We have that

\[\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}n^k=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} \sum_{n=1}^\infty n^k\]

So far, the only thing I've done that is illegal is swapping the summation order of k and n (but I personally feel that swapping summation is always valid when viewed in the right light). Now, the next step will be doing something very illegal-- replacing the divergent sum \( \sum_{n=1}^\infty n^k \) with \( \zeta(-k) \). Now, if you read the MO post, you know that making this replacement requires picking up the residue of \( \zeta(-k) \). Before, we do that, lets look at what the sum has become. We now have (using the relationship between the zeta function and bernoulli numbers)

\[\sum f(n) = \sum \frac{f^{k}(0)}{k!} \zeta(-k) = -\sum \frac{f^{(k)}(0)}{k!} \frac{B_{k+1}}{k+1} = -\sum_{k=0}^\infty \frac{f^{(2k-1)}(0)}{(2k)!} B_{2k}\]

WOW! This is exactly the Euler-Maclaurn Formula-- except with one missing term, the integral. This missing term is precisely what is picked up by the residue of the \( \zeta \) function. In particular, we can rewrite out sum as the contour integral

\[\int_{c-i \infty}^{c + i \infty} \frac{1}{e^{2 \pi i k}-1}\frac{f^{(k)}(0)}{k!} \zeta(-k) dk\]

Lo and behold, there is an extra residue at -1, with value exactly equal to \( f^{(-1)}(0) = \int_0^\infty f(x)dx \). Therefore, my approach allows us to easily obtain the missing term that can be missed in a usual divergent series approach to looking at the E-M formula. I don't go over this case in too much detail, since I've already written about it HERE.

Links to some other cases:

I've used this trick in a few other instances to easily obtain results about analytical continuation. For instance, it provides an easy way to figure out that tommy's zeta function has a branch cut See here. Another similar example can be shown to have a branch cut using this same approach See here.

Case 2; The main study: \( \sum (-1)^n x^{2^n} \)

This is a very trick series to work with (if you want more info about other approachs see here). However, using my approach, we get a mind-numbingly easy way to sum this series. First, do the following simplification

\[ \sum (-1)^n e^{\ln(x) 2^n} = \sum_n (-1)^n \sum_k \frac{\ln(x)^k 2^{nk}}{k!} = \sum_k \frac{\ln(x)^k}{k!}\sum_n (-1)^n 2^{nk}\]

Now, we do the trick where we replace the \(\sum (-1)^n 2^{nk}\) by its analytical continuation, which is \(\frac{1}{1+2^n}\). Now, notice that we introduce a residue at each \( 1+ 2^z = 0 \implies z = \frac{-\pi i + 2 \pi i n}{\ln(2)} \). Also, notice that the residue at \(\frac{1}{1+2^z}\) has a value of \(\frac{2 \pi i}{\ln(2)}\)

Therefore, the sum becomes the contour integral

\[ \int_{-1/2 - i \infty}^{-1/2 + i \infty} \frac{\ln(x)^k }{k!}\frac{1}{1+2^k}\frac{1}{e^{2 \pi i k}-1} dk\]

or, written as a sum, and letting \( \overline n = \frac{-\pi i + 2 \pi i n}{\ln(2)} \) then

\[ \sum\frac{\ln(x)^k}{k!} \frac{1}{1+2^k} + \sum_{n=-\infty}^\infty \frac{2 \pi i}{\ln(2)} \frac{\ln(x)^{\overline{n}}}{(\overline{n})!} \frac{1}{e^{2 \pi i \overline{n}}-1} \]

There's also another interesting aspect of this approach. Take the sum \( \sum (-1)^n x^{a^n} \). If we have \( |a|<1 \) then we actually have that the inner sum \( \sum (-1)^n a^{nk}\) doesn't diverge. That means that we shouldn't pick up the residues. And in fact, we don't need to! In fact, we have the interesting relationship that

\[\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2}\right)^{z^n}-\sum_{k=0}^\infty \frac{\ln(\frac{1}{2})^k}{k!(1+z^k)} =

\begin{cases}

0 & |z|<1 \\

\sum \text{Res}(f,a_k) & |z|>1 \\

\end{cases}\]

Where RES is supposed to be the sum over the extra residues that get created (its a bit tedious to write out, but hopefully you get the idea).

Problems and Analysis of the General Case

I've been using this approach for a while, and it tended to only give correct answers, so I assume it works in a pretty wide array of situations, but I think I'm still missing some theory on when I'm supposed to pick up residues. Unforunately, I don't possess a great deal of mathematical skill, so I will proceed to try to study the general case by looking at a large number of examples. In particular, I will look at studying the relationship between the following pairs of series. (Also, I would be grateful to see anyone's ideas on how to unify these ideas in a mathemtically rigourous, so I don't have to resort to lots of examples to illustrate my points).

Also, I quick note of notation, I have \( F^P(x) = \sum_{n=0}^\infty f(n)x^n \), \( F^D(x) = \sum_{n=1}^\infty \frac{f(n)}{n^x} \). Where the P is for 'Power Series' and the D for 'Dirichlet Series.'

Let's consider the simplest possible example, where \( f(k) =1 \). Then we have that \( F^D(-zn) = \zeta(-zn) \). Unforunately, we don't have that either of the series converges at the same time. Also, note that I think the LHS series actually has a natural boundary on the line \( \mathfrak{Re}(z) = 0 \). So, in order to study this in a reasonable way, we will need to add in a factor so that \( \sum_{k=1}^\infty \frac{f(k)}{1+k^z} \) converges everywhere. When this sum converges everywhere, I will refer to it as the canonical extension of \( \sum_{k=1}^\infty \frac{f(k)}{1+k^z} \). I call it an extension because if I give the \( \sum_{k=1}^\infty \frac{f(k)}{1+k^z} \) for \( \mathfrak{Re}(z) > 0 \) then we can't analytically continue this function to \( \mathfrak{Re}(z) < 0 \). However, the most natural extension is to simply evaulate the sum on the other side of the natural boundary, and so I call this natural-seeming extension that canonical extension. I'll come back to \( f(k) = 1 \) later, but for now, I'll study the easier case of \( f(k) =\frac{1}{k^2} \), so that the series \( \sum_{k=1}^\infty \frac{f(k)}{1+k^z} \) converges absolutely and everywhere.

So, lets do some algebra. We have that

\[ (\star_1) \quad \sum_{k=1}^\infty \frac{1}{k^2}\frac{1}{1+k^z} = \sum_{k=1}^\infty \frac{1}{k^2} \sum_{n=0}^\infty (-1)^n k^{zn} \]

\[\sum_{k=1}^\infty \frac{1}{k^2} \sum_{n=0}^\infty (-1)^n k^{zn} = \sum_{n=0}^\infty(-1)^n \sum_{k=1}^\infty k^{zn-2}\]

\[(\star_2) \quad \sum_{n=0}^\infty (-1)^n \sum_{k=1}^\infty k^{zn-2} = \sum_{n=0}^\infty (-1)^n \zeta(-zn+2)\]

In both \(\star_1\) and \(\star_2\) I preform an illegal step where I replace a divergent series by its analaytical continuation. If my theory were correct, then in both steps we need to pick up the residues. However, it appears only \( \star_2 \) creates any new residues. Note that we only need to pick up a residue when \( \mathfrak{Re}(z)> 0 \), since otherwise the the Dirichlet series is not divergent. Doing some computations, we get that

\[\sum_{k=1}^\infty \frac{1}{k^2}\frac{1}{1+k^z} - \sum_{n=0}^\infty (-1)^n \zeta(-zn+2) = 0\]

When \( \mathfrak{Re}(z) < 0 \). Note that the series on the RHS doesn't actually quite converge, since it alternates between 1 and -1, but if we do cesaro summation or anything else of the same sort we get that the sums are equal. However, when \( \mathfrak{Re}(z) > 0 \) we get that their difference is exactly equal to the residues introduced by \( \zeta(-zn+2)\). Note that there is a residue at \( n = 1/z\) and \( n = + \infty \). As an aside, that residue at \( +\infty \) is a bit harder to compute directly than a regular residue. But, there's an easy way to see that its there (and to compute its contribution). If we take z as an even integer \( z=2k \), then \( \zeta(-(2k)n + 2) = 0\). So, the contour integral

\[ \int_{c - i \infty}^{c + i \infty} \zeta(-(2k)n + 2) \csc(\pi n) dn \]

Doesn't enclose any residues (for c sufficently large), so it should be zero-- right? Well actually, the contour integral doesn't evaluate to zero, and that's because the contour integral picks up the residue at \( +\infty\) which is non-zero.

Thus, for \( \mathfrak{Re}(z) > 0 \), we actually have

\[ \sum_{k=1}^\infty \frac{1}{k^2}\frac{1}{1+k^z} - \sum_{n=0}^\infty (-1)^n \zeta(-zn+2) \neq 0\]

And instead we have

\[\sum_{k=1}^\infty \frac{1}{k^2}\frac{1}{1+k^z} = \int_C \zeta(-zn+2) \frac{\csc(\pi n)}{2i} dn\]

Where C is a contour that encloses the residue of \(\zeta\) at \( n = \frac{1}{z} \) and also picks up the residue at infinity. Since \(\frac{1}{z} > 0\), its very easy to pick make a contour, just choose something like

\[\int_{c - i \infty}^{c + i \infty} \zeta(-zn+2) \frac{\csc(\pi n)}{2i}dn \]

for \(-1<c< 0\).

Another example is given by choosing \( f(k) = \frac{\mu(k)}{k^2} \) Now, preforming the same steps as before we get that

\[ (\star_1) \quad \sum_{k=1}^\infty \frac{\mu(k)}{k^2}\frac{1}{1+k^z} = \sum_{k=1}^\infty \frac{\mu(k)}{k^2} \sum_{n=0}^\infty (-1)^n k^{zn} \]

\[\sum_{k=1}^\infty \frac{\mu(k)}{k^2} \sum_{n=0}^\infty (-1)^n k^{zn} = \sum_{n=0}^\infty(-1)^n \sum_{k=1}^\infty \mu(k) k^{zn-2}\]

\[(\star_2) \quad \sum_{n=0}^\infty (-1)^n \sum_{k=1}^\infty \mu(k)k^{zn-2} = \sum_{n=0}^\infty (-1)^n \frac{1}{\zeta(-zn+2)}\]

Again, we see that \( \star_1 \) doesn't contribute any residues, but \( \star_2 \) does. In this case, there is no residue at \(\infty\) which is nice. This means we can more easily compute the difference in a 'closed form' kind of way. We have that

\[\sum_{k=1}^\infty \frac{\mu(k)}{k^2}\frac{1}{1+k^z} - \sum_{n=0}^\infty (-1)^n \frac{1}{\zeta(-zn+2)} = 0\]

When \( \mathfrak{Re}(z) < 0 \). (Note that we again have to apply Cesaro summation to get the RHS to converge). However, when \( \mathfrak{Re}(z) > 0 \) we need to pick up all the zeroes of the \(\zeta\) function. We have a residue at \(-zn +2 = \rho \implies n=\frac{\rho-2}{-z}\). Thus, the difference becomes

\[\sum_{k=1}^\infty \frac{\mu(k)}{k^2}\frac{1}{1+k^z} - \sum_{n=0}^\infty (-1)^n \frac{1}{\zeta(-zn+2)} = \sum_{p = \frac{2-\rho}{z}} \text{Res}\left( \frac{\csc(\pi p)}{2i \zeta(-zp+2)}\right)\]

An easier way to compute this is with a contour integral, which will pick up everything for us, so we have

\[\sum_{k=1}^\infty \frac{\mu(k)}{k^2}\frac{1}{1+k^z} - \sum_{n=0}^\infty (-1)^n \frac{1}{\zeta(-zn+2)} = \int_{-1/2 - i \infty}^{-1/2 + i \infty} \frac{\csc(\pi n)}{2i \zeta(-zn+2)} dn \]

I've also tested out this method for \( f(k) = \frac{\lambda(k)}{k^2} \), and I obtain the same sort of results.

The result seems to be that \( \star_1 \) doesn't contribute any residues, but \( \star_2 \) generally still does. Thus, apparently, \( \frac{1}{1+k^z}\) doesn't appear to create extra residues.

The upshot is that we can study the sum of the poles of \( F^D(-zn) \) by studying

\( \sum_{k=1}^\infty \frac{f(k)}{1+k^z} - \sum_{n=0}^\infty (F^D(-zn)) \). I imagine that with a suitable choice of functions, we could probably reduce the Riemann hypothesis to some statement about the difference between the sums-- which is interesting, since we are getting some number theoretic statements out of studying functions beyond their natural boundaries (of course though, one might expect that this just makes the statement harder than it was orginally).

(2)

To get directly to the punch line, it appears that if we look at \( \frac{1}{1+z^k}\), then it appears to contribute residues always. This is, of course, in strong constrast to the last example. Now, lets look at some examples.

We have already explored the case where \( f(k) = \frac{1}{k!} \), and seen that in this case residues get picked up along the imaginary axis when \( |z| > 1 \).

Let us consider a closely related sum \( f(k) = \frac{\sin(\frac{\pi}{2} k)}{k!} \), then \( F^P(k) = \sin(k) \). Lets run roughly the same argument we gave in part 1.

\[ (\star_1) \quad \sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+z^k} =\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\sum_{n=0}^\infty (-1)^n z^{kn}\]

\[\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\sum_{n=0}^\infty (-1)^n z^{kn} = \sum_{n=0}^\infty (-1)^n \sin(z^n) \]

Now, notice that in the second step the sum was convergent, and describes a holomorphic function, so the only possible source of extra residues is from \( \star_1 \). We obtain that

\[\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+z^k} = \sum_{n=0}^\infty (-1)^n \sin(z^n)\]

For \( |z|<1 \). For \( |z|>1 \), as we expect, there are residues that are picked up. Using a contour integral again, we have that

\[\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+z^k} =\int_{-1/2 - i \infty}^{-1/2 + i \infty} \frac{\csc(\pi n)}{2i} \sin(z^n) dn\]

Now, lets look at a very closely related series, we instead look at

\[\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+\frac{z^k}{2}} \sim \sum_{n=0}^\infty (-1)^n \frac{\sin(z^n)}{2^n}\]

The RHS creates a pretty weird function (I wrote a low quality reddit post about it here)

But, with this change, now both sides converge, thus we can talk about

\[\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+\frac{z^k}{2}}- \sum_{n=0}^\infty (-1)^n \frac{\sin(z^n)}{2^n}\]

Which of course is zero for \( |z|<1 \). We could do the same contour integral approach to find the difference. However, there is a better, and far more interesting way. At the \( \star_1 \) step, we could choose to either pick up there residues of the function on the RHS, or we could pick up the residues of the LHS, which are at \(\frac{1}{1+\frac{z^k}{2}} = \infty\). Doing this, the extra residues are at \( 1+\frac{z^k}{2} = 0 \implies \overline{k} = \frac{\ln(2) + \pi i + 2\pi i m}{\ln(z)}, m \in \mathbb{Z} \)

Now, the important part is this: the location of the residues are on the imaginary axis!!! This tells us that summation beyond natural boundary's is deeply connected to extensions of coefficents into the complex plane.

In particular, we obtain the relationship that

\[ \sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+\frac{z^k}{2}} + \sum_{\overline{k}}\frac{f(\overline{k})}{e^{2 \pi i \overline{k}}}\text{Res}(\frac{1}{1+z^k/2}, k = \overline{k}) = \sum_{n=0}^\infty (-1)^n \frac{\sin(z^n)}{2^n} \]

for \(|z|>1 \). Well, actually, as written it won't converge, so instead we need to the second sum it in a slightly different way. This series gives something that is cesaro summable

\[\sum_{\overline{k}} \frac{\sin(\pi/2 \overline{k}) e^{\pi i \overline{k}}}{e^{2 \pi i \overline{k}}}\frac{1}{\overline{k}!}\frac{1}{e^{2 \pi i \overline{k}}}\text{Res}(\frac{1}{1+z^k/2}, k = \overline{k})\]

Both of the previous examples were pretty closely connected with the gamma function, so let me next pick an example where there is more than just the \( \Gamma \) function involved. I will use the fact that \( \frac{-z e^{-z}}{e^{-z} - 1} = \sum_{n=0}^\infty (-1)^n \frac{B^{+}_n}{n!} z^n \) where \( B^{+}_n = -n \zeta(1-n) \) are the Bernoulli numbers. Then we have that

\[\sum_{k=0}^\infty \frac{-k \zeta(1-k)(-1)^k }{k!} \frac{1}{1+\frac{z^k}{2}} \sim \sum_{n=0}^\infty \frac{(-1)^n}{2^n} \frac{-z^n e^{-z^n}}{e^{-z^n}-1} \]

These two series are equal for \( |z|<1 \). The difference when \( |z|>1 \) becomes equal to the extra residues of produced by \(\frac{1}{1+z^k/2}\). We can write this difference between the two functions precisely as

\[\sum_{\overline{k}} \frac{\csc(\pi \overline{k})}{2i} \frac{-\overline{k} \zeta(1-\overline{k})(-1)^\overline{k} }{\overline{k}!}\text{Res}(\frac{1}{1+z^k/2}, k = \overline{k})\]

So, the difference now depends on the zeta function on at imaginary values. Again, I should emphasize, we started by looking at the Bernoulli numbers-- which only make sense at non-negative integers. But, the solution required us to look at those coefficents at imaginary values.

READER'S NOTE: I plan to add in sections (3) and (4) later. My hope is that they might provide some framework for continuing coefficients defined on integers into the imaginary values. However, it will be a bit of time before I can go back and write those sections, so I'm posting what I have for now.

Some thoughts

I think there is some very fascinating and deep going on here. This are some preliminary thoughts on how I think about it.

First, observe that continuation beyond natural boundaries is not unique. There are many different extensions we could choose. Also, observe that extensions of coefficent functions (I'll denote it \( a_n \) defined on the integers into the complex plane is not unique. I think these two facts are linked together in the following way. Section (2) involved taking an extension of an analytic function (call it \( F(z) \) beyond a natural boundary. For each extension of \( F(z) \) we picked, we essentially induce a definition of \( a_n \) on the complex plane. Therefore, if there exists a canonical extension of \( F(z) \), then we should induce a canonical extension of \( a_n \) into the plane. We can also go the other direction-- by studying what canonical extensions of \( a_n \) look like in general, we could perhaps derive a canonical extension of a given \( a_n \) and use this to induce an extension of \( F(z) \).

In particular, I think finding an canonical extension of the Liouville \( \lambda \) function onto the complex plane would provide a canonical extension of the Jacobi theta function. This opens a number of deeply fascinating directions of study. For instance, can we study modular forms by studying number theoretic functions extended into the complex plane? I know one implication of this method would be that we can study the primes by looking at extensions of the Moebius function into the complex plane, since that would allow us to study the prime zeta function beyond its natural boundary. There are lots and lots of interesting number-theoretic identities involving Lambert Series, which is basically the form studied in (2). Thus, it conceivable that being able to understand deeply whats going on in (2) could open lots of door for studying number theory. I think a good place to start for studying this would be to look at the Mangolt function. It has a particularly simple relationship \( \sum \Lambda(n) \frac{q^n}{1-q^n} = \sum \ln(n) q^n \). The RHS can be analytically continued using the Lerch Phi function, and this continuation might provide a way to go backwards an induce a value for \( \Lambda \). Probably having a specific case like this to study will be helpful.

Anyway-- thanks for getting all the way down to this point in the post. I'm guessing you probably have lots of questions, and there are probably a few mistakes in my math in the space above-- though I tried to closely check that everything numerically checks out at the least-- so please ask for clarification anywhere that seems confusing or wrong so I can improve my post. Also, any thoughts or insights are appreciated.

Thanks for reading!

Also, I recommend you read this MO post for context, since it contains lots of background and context for this question.

Since this is a long post, I assume most people won't want to read the whole thing-- if your willing to grant that the method works fairly generally and is useful, I'd recommend

- Skim up to 'Some thoughts' on https://mathoverflow.net/questions/43869...tical-cont

- Skim through one example in section (1) and (2) and read the bolded stuff

- Read the final thoughts at the end of section (1) and (2)

Background

An enormously useful trick in analyzing an infinite series of the form \( \sum_{n=1}^\infty f(n) \) is to expand \( f(n) \) as its own infinite series, and then swap the order of summation, and sum the inner series by analytical continuation. If you read the linked MO post, then you saw my example where the naive approach of simply replacing the inner series by analytical continuation fails. My solution to this problem was to pick up the residues of the analytical continuation of the inner series. In doing some recent calculations, I've come to realize that my solution doesn't quite work-- but I think its definitely on the right track. Thus, to start, let me illustrate some cases where my theory produces some useful results.

Case 1: Euler-Maclaurin Formula

Let us assume that \( f(n) \) can be represented by a power series at 0. Then \(f(n) = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} n^k \). Now, lets try to analyze the series \( \sum_{n=1}^\infty f(n)\). We have that

\[\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}n^k=\sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} \sum_{n=1}^\infty n^k\]

So far, the only thing I've done that is illegal is swapping the summation order of k and n (but I personally feel that swapping summation is always valid when viewed in the right light). Now, the next step will be doing something very illegal-- replacing the divergent sum \( \sum_{n=1}^\infty n^k \) with \( \zeta(-k) \). Now, if you read the MO post, you know that making this replacement requires picking up the residue of \( \zeta(-k) \). Before, we do that, lets look at what the sum has become. We now have (using the relationship between the zeta function and bernoulli numbers)

\[\sum f(n) = \sum \frac{f^{k}(0)}{k!} \zeta(-k) = -\sum \frac{f^{(k)}(0)}{k!} \frac{B_{k+1}}{k+1} = -\sum_{k=0}^\infty \frac{f^{(2k-1)}(0)}{(2k)!} B_{2k}\]

WOW! This is exactly the Euler-Maclaurn Formula-- except with one missing term, the integral. This missing term is precisely what is picked up by the residue of the \( \zeta \) function. In particular, we can rewrite out sum as the contour integral

\[\int_{c-i \infty}^{c + i \infty} \frac{1}{e^{2 \pi i k}-1}\frac{f^{(k)}(0)}{k!} \zeta(-k) dk\]

Lo and behold, there is an extra residue at -1, with value exactly equal to \( f^{(-1)}(0) = \int_0^\infty f(x)dx \). Therefore, my approach allows us to easily obtain the missing term that can be missed in a usual divergent series approach to looking at the E-M formula. I don't go over this case in too much detail, since I've already written about it HERE.

Links to some other cases:

I've used this trick in a few other instances to easily obtain results about analytical continuation. For instance, it provides an easy way to figure out that tommy's zeta function has a branch cut See here. Another similar example can be shown to have a branch cut using this same approach See here.

Case 2; The main study: \( \sum (-1)^n x^{2^n} \)

This is a very trick series to work with (if you want more info about other approachs see here). However, using my approach, we get a mind-numbingly easy way to sum this series. First, do the following simplification

\[ \sum (-1)^n e^{\ln(x) 2^n} = \sum_n (-1)^n \sum_k \frac{\ln(x)^k 2^{nk}}{k!} = \sum_k \frac{\ln(x)^k}{k!}\sum_n (-1)^n 2^{nk}\]

Now, we do the trick where we replace the \(\sum (-1)^n 2^{nk}\) by its analytical continuation, which is \(\frac{1}{1+2^n}\). Now, notice that we introduce a residue at each \( 1+ 2^z = 0 \implies z = \frac{-\pi i + 2 \pi i n}{\ln(2)} \). Also, notice that the residue at \(\frac{1}{1+2^z}\) has a value of \(\frac{2 \pi i}{\ln(2)}\)

Therefore, the sum becomes the contour integral

\[ \int_{-1/2 - i \infty}^{-1/2 + i \infty} \frac{\ln(x)^k }{k!}\frac{1}{1+2^k}\frac{1}{e^{2 \pi i k}-1} dk\]

or, written as a sum, and letting \( \overline n = \frac{-\pi i + 2 \pi i n}{\ln(2)} \) then

\[ \sum\frac{\ln(x)^k}{k!} \frac{1}{1+2^k} + \sum_{n=-\infty}^\infty \frac{2 \pi i}{\ln(2)} \frac{\ln(x)^{\overline{n}}}{(\overline{n})!} \frac{1}{e^{2 \pi i \overline{n}}-1} \]

There's also another interesting aspect of this approach. Take the sum \( \sum (-1)^n x^{a^n} \). If we have \( |a|<1 \) then we actually have that the inner sum \( \sum (-1)^n a^{nk}\) doesn't diverge. That means that we shouldn't pick up the residues. And in fact, we don't need to! In fact, we have the interesting relationship that

\[\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2}\right)^{z^n}-\sum_{k=0}^\infty \frac{\ln(\frac{1}{2})^k}{k!(1+z^k)} =

\begin{cases}

0 & |z|<1 \\

\sum \text{Res}(f,a_k) & |z|>1 \\

\end{cases}\]

Where RES is supposed to be the sum over the extra residues that get created (its a bit tedious to write out, but hopefully you get the idea).

Problems and Analysis of the General Case

I've been using this approach for a while, and it tended to only give correct answers, so I assume it works in a pretty wide array of situations, but I think I'm still missing some theory on when I'm supposed to pick up residues. Unforunately, I don't possess a great deal of mathematical skill, so I will proceed to try to study the general case by looking at a large number of examples. In particular, I will look at studying the relationship between the following pairs of series. (Also, I would be grateful to see anyone's ideas on how to unify these ideas in a mathemtically rigourous, so I don't have to resort to lots of examples to illustrate my points).

Also, I quick note of notation, I have \( F^P(x) = \sum_{n=0}^\infty f(n)x^n \), \( F^D(x) = \sum_{n=1}^\infty \frac{f(n)}{n^x} \). Where the P is for 'Power Series' and the D for 'Dirichlet Series.'

- \[ \sum_{k=1}^\infty \frac{f(k)}{1+k^z} \sim \sum_{n=0}^\infty (F^D(-zn)) \]

- \[ \sum_{k=1}^\infty \frac{f(k)}{1-z^k} \sim \sum_{n=0}^\infty (F^P(z^n)-f(0))\]

- \[ \sum_{k=0}^\infty \frac{f(k)}{z-k} \sim \sum_{n=0}^\infty \frac{F^D(-n)}{z^{n+1}} \]

- \[ \sum_{k=0}^\infty f(k) \zeta(-zk) \sim \sum_{n=1}^\infty F^P(n^z) \]

Let's consider the simplest possible example, where \( f(k) =1 \). Then we have that \( F^D(-zn) = \zeta(-zn) \). Unforunately, we don't have that either of the series converges at the same time. Also, note that I think the LHS series actually has a natural boundary on the line \( \mathfrak{Re}(z) = 0 \). So, in order to study this in a reasonable way, we will need to add in a factor so that \( \sum_{k=1}^\infty \frac{f(k)}{1+k^z} \) converges everywhere. When this sum converges everywhere, I will refer to it as the canonical extension of \( \sum_{k=1}^\infty \frac{f(k)}{1+k^z} \). I call it an extension because if I give the \( \sum_{k=1}^\infty \frac{f(k)}{1+k^z} \) for \( \mathfrak{Re}(z) > 0 \) then we can't analytically continue this function to \( \mathfrak{Re}(z) < 0 \). However, the most natural extension is to simply evaulate the sum on the other side of the natural boundary, and so I call this natural-seeming extension that canonical extension. I'll come back to \( f(k) = 1 \) later, but for now, I'll study the easier case of \( f(k) =\frac{1}{k^2} \), so that the series \( \sum_{k=1}^\infty \frac{f(k)}{1+k^z} \) converges absolutely and everywhere.

So, lets do some algebra. We have that

\[ (\star_1) \quad \sum_{k=1}^\infty \frac{1}{k^2}\frac{1}{1+k^z} = \sum_{k=1}^\infty \frac{1}{k^2} \sum_{n=0}^\infty (-1)^n k^{zn} \]

\[\sum_{k=1}^\infty \frac{1}{k^2} \sum_{n=0}^\infty (-1)^n k^{zn} = \sum_{n=0}^\infty(-1)^n \sum_{k=1}^\infty k^{zn-2}\]

\[(\star_2) \quad \sum_{n=0}^\infty (-1)^n \sum_{k=1}^\infty k^{zn-2} = \sum_{n=0}^\infty (-1)^n \zeta(-zn+2)\]

In both \(\star_1\) and \(\star_2\) I preform an illegal step where I replace a divergent series by its analaytical continuation. If my theory were correct, then in both steps we need to pick up the residues. However, it appears only \( \star_2 \) creates any new residues. Note that we only need to pick up a residue when \( \mathfrak{Re}(z)> 0 \), since otherwise the the Dirichlet series is not divergent. Doing some computations, we get that

\[\sum_{k=1}^\infty \frac{1}{k^2}\frac{1}{1+k^z} - \sum_{n=0}^\infty (-1)^n \zeta(-zn+2) = 0\]

When \( \mathfrak{Re}(z) < 0 \). Note that the series on the RHS doesn't actually quite converge, since it alternates between 1 and -1, but if we do cesaro summation or anything else of the same sort we get that the sums are equal. However, when \( \mathfrak{Re}(z) > 0 \) we get that their difference is exactly equal to the residues introduced by \( \zeta(-zn+2)\). Note that there is a residue at \( n = 1/z\) and \( n = + \infty \). As an aside, that residue at \( +\infty \) is a bit harder to compute directly than a regular residue. But, there's an easy way to see that its there (and to compute its contribution). If we take z as an even integer \( z=2k \), then \( \zeta(-(2k)n + 2) = 0\). So, the contour integral

\[ \int_{c - i \infty}^{c + i \infty} \zeta(-(2k)n + 2) \csc(\pi n) dn \]

Doesn't enclose any residues (for c sufficently large), so it should be zero-- right? Well actually, the contour integral doesn't evaluate to zero, and that's because the contour integral picks up the residue at \( +\infty\) which is non-zero.

Thus, for \( \mathfrak{Re}(z) > 0 \), we actually have

\[ \sum_{k=1}^\infty \frac{1}{k^2}\frac{1}{1+k^z} - \sum_{n=0}^\infty (-1)^n \zeta(-zn+2) \neq 0\]

And instead we have

\[\sum_{k=1}^\infty \frac{1}{k^2}\frac{1}{1+k^z} = \int_C \zeta(-zn+2) \frac{\csc(\pi n)}{2i} dn\]

Where C is a contour that encloses the residue of \(\zeta\) at \( n = \frac{1}{z} \) and also picks up the residue at infinity. Since \(\frac{1}{z} > 0\), its very easy to pick make a contour, just choose something like

\[\int_{c - i \infty}^{c + i \infty} \zeta(-zn+2) \frac{\csc(\pi n)}{2i}dn \]

for \(-1<c< 0\).

Another example is given by choosing \( f(k) = \frac{\mu(k)}{k^2} \) Now, preforming the same steps as before we get that

\[ (\star_1) \quad \sum_{k=1}^\infty \frac{\mu(k)}{k^2}\frac{1}{1+k^z} = \sum_{k=1}^\infty \frac{\mu(k)}{k^2} \sum_{n=0}^\infty (-1)^n k^{zn} \]

\[\sum_{k=1}^\infty \frac{\mu(k)}{k^2} \sum_{n=0}^\infty (-1)^n k^{zn} = \sum_{n=0}^\infty(-1)^n \sum_{k=1}^\infty \mu(k) k^{zn-2}\]

\[(\star_2) \quad \sum_{n=0}^\infty (-1)^n \sum_{k=1}^\infty \mu(k)k^{zn-2} = \sum_{n=0}^\infty (-1)^n \frac{1}{\zeta(-zn+2)}\]

Again, we see that \( \star_1 \) doesn't contribute any residues, but \( \star_2 \) does. In this case, there is no residue at \(\infty\) which is nice. This means we can more easily compute the difference in a 'closed form' kind of way. We have that

\[\sum_{k=1}^\infty \frac{\mu(k)}{k^2}\frac{1}{1+k^z} - \sum_{n=0}^\infty (-1)^n \frac{1}{\zeta(-zn+2)} = 0\]

When \( \mathfrak{Re}(z) < 0 \). (Note that we again have to apply Cesaro summation to get the RHS to converge). However, when \( \mathfrak{Re}(z) > 0 \) we need to pick up all the zeroes of the \(\zeta\) function. We have a residue at \(-zn +2 = \rho \implies n=\frac{\rho-2}{-z}\). Thus, the difference becomes

\[\sum_{k=1}^\infty \frac{\mu(k)}{k^2}\frac{1}{1+k^z} - \sum_{n=0}^\infty (-1)^n \frac{1}{\zeta(-zn+2)} = \sum_{p = \frac{2-\rho}{z}} \text{Res}\left( \frac{\csc(\pi p)}{2i \zeta(-zp+2)}\right)\]

An easier way to compute this is with a contour integral, which will pick up everything for us, so we have

\[\sum_{k=1}^\infty \frac{\mu(k)}{k^2}\frac{1}{1+k^z} - \sum_{n=0}^\infty (-1)^n \frac{1}{\zeta(-zn+2)} = \int_{-1/2 - i \infty}^{-1/2 + i \infty} \frac{\csc(\pi n)}{2i \zeta(-zn+2)} dn \]

I've also tested out this method for \( f(k) = \frac{\lambda(k)}{k^2} \), and I obtain the same sort of results.

The result seems to be that \( \star_1 \) doesn't contribute any residues, but \( \star_2 \) generally still does. Thus, apparently, \( \frac{1}{1+k^z}\) doesn't appear to create extra residues.

The upshot is that we can study the sum of the poles of \( F^D(-zn) \) by studying

\( \sum_{k=1}^\infty \frac{f(k)}{1+k^z} - \sum_{n=0}^\infty (F^D(-zn)) \). I imagine that with a suitable choice of functions, we could probably reduce the Riemann hypothesis to some statement about the difference between the sums-- which is interesting, since we are getting some number theoretic statements out of studying functions beyond their natural boundaries (of course though, one might expect that this just makes the statement harder than it was orginally).

(2)

To get directly to the punch line, it appears that if we look at \( \frac{1}{1+z^k}\), then it appears to contribute residues always. This is, of course, in strong constrast to the last example. Now, lets look at some examples.

We have already explored the case where \( f(k) = \frac{1}{k!} \), and seen that in this case residues get picked up along the imaginary axis when \( |z| > 1 \).

Let us consider a closely related sum \( f(k) = \frac{\sin(\frac{\pi}{2} k)}{k!} \), then \( F^P(k) = \sin(k) \). Lets run roughly the same argument we gave in part 1.

\[ (\star_1) \quad \sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+z^k} =\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\sum_{n=0}^\infty (-1)^n z^{kn}\]

\[\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\sum_{n=0}^\infty (-1)^n z^{kn} = \sum_{n=0}^\infty (-1)^n \sin(z^n) \]

Now, notice that in the second step the sum was convergent, and describes a holomorphic function, so the only possible source of extra residues is from \( \star_1 \). We obtain that

\[\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+z^k} = \sum_{n=0}^\infty (-1)^n \sin(z^n)\]

For \( |z|<1 \). For \( |z|>1 \), as we expect, there are residues that are picked up. Using a contour integral again, we have that

\[\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+z^k} =\int_{-1/2 - i \infty}^{-1/2 + i \infty} \frac{\csc(\pi n)}{2i} \sin(z^n) dn\]

Now, lets look at a very closely related series, we instead look at

\[\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+\frac{z^k}{2}} \sim \sum_{n=0}^\infty (-1)^n \frac{\sin(z^n)}{2^n}\]

The RHS creates a pretty weird function (I wrote a low quality reddit post about it here)

But, with this change, now both sides converge, thus we can talk about

\[\sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+\frac{z^k}{2}}- \sum_{n=0}^\infty (-1)^n \frac{\sin(z^n)}{2^n}\]

Which of course is zero for \( |z|<1 \). We could do the same contour integral approach to find the difference. However, there is a better, and far more interesting way. At the \( \star_1 \) step, we could choose to either pick up there residues of the function on the RHS, or we could pick up the residues of the LHS, which are at \(\frac{1}{1+\frac{z^k}{2}} = \infty\). Doing this, the extra residues are at \( 1+\frac{z^k}{2} = 0 \implies \overline{k} = \frac{\ln(2) + \pi i + 2\pi i m}{\ln(z)}, m \in \mathbb{Z} \)

Now, the important part is this: the location of the residues are on the imaginary axis!!! This tells us that summation beyond natural boundary's is deeply connected to extensions of coefficents into the complex plane.

In particular, we obtain the relationship that

\[ \sum_{k=0}^\infty \frac{\sin(\frac{\pi}{2} k)}{k!}\frac{1}{1+\frac{z^k}{2}} + \sum_{\overline{k}}\frac{f(\overline{k})}{e^{2 \pi i \overline{k}}}\text{Res}(\frac{1}{1+z^k/2}, k = \overline{k}) = \sum_{n=0}^\infty (-1)^n \frac{\sin(z^n)}{2^n} \]

for \(|z|>1 \). Well, actually, as written it won't converge, so instead we need to the second sum it in a slightly different way. This series gives something that is cesaro summable

\[\sum_{\overline{k}} \frac{\sin(\pi/2 \overline{k}) e^{\pi i \overline{k}}}{e^{2 \pi i \overline{k}}}\frac{1}{\overline{k}!}\frac{1}{e^{2 \pi i \overline{k}}}\text{Res}(\frac{1}{1+z^k/2}, k = \overline{k})\]

Both of the previous examples were pretty closely connected with the gamma function, so let me next pick an example where there is more than just the \( \Gamma \) function involved. I will use the fact that \( \frac{-z e^{-z}}{e^{-z} - 1} = \sum_{n=0}^\infty (-1)^n \frac{B^{+}_n}{n!} z^n \) where \( B^{+}_n = -n \zeta(1-n) \) are the Bernoulli numbers. Then we have that

\[\sum_{k=0}^\infty \frac{-k \zeta(1-k)(-1)^k }{k!} \frac{1}{1+\frac{z^k}{2}} \sim \sum_{n=0}^\infty \frac{(-1)^n}{2^n} \frac{-z^n e^{-z^n}}{e^{-z^n}-1} \]

These two series are equal for \( |z|<1 \). The difference when \( |z|>1 \) becomes equal to the extra residues of produced by \(\frac{1}{1+z^k/2}\). We can write this difference between the two functions precisely as

\[\sum_{\overline{k}} \frac{\csc(\pi \overline{k})}{2i} \frac{-\overline{k} \zeta(1-\overline{k})(-1)^\overline{k} }{\overline{k}!}\text{Res}(\frac{1}{1+z^k/2}, k = \overline{k})\]

So, the difference now depends on the zeta function on at imaginary values. Again, I should emphasize, we started by looking at the Bernoulli numbers-- which only make sense at non-negative integers. But, the solution required us to look at those coefficents at imaginary values.

READER'S NOTE: I plan to add in sections (3) and (4) later. My hope is that they might provide some framework for continuing coefficients defined on integers into the imaginary values. However, it will be a bit of time before I can go back and write those sections, so I'm posting what I have for now.

Some thoughts

I think there is some very fascinating and deep going on here. This are some preliminary thoughts on how I think about it.

First, observe that continuation beyond natural boundaries is not unique. There are many different extensions we could choose. Also, observe that extensions of coefficent functions (I'll denote it \( a_n \) defined on the integers into the complex plane is not unique. I think these two facts are linked together in the following way. Section (2) involved taking an extension of an analytic function (call it \( F(z) \) beyond a natural boundary. For each extension of \( F(z) \) we picked, we essentially induce a definition of \( a_n \) on the complex plane. Therefore, if there exists a canonical extension of \( F(z) \), then we should induce a canonical extension of \( a_n \) into the plane. We can also go the other direction-- by studying what canonical extensions of \( a_n \) look like in general, we could perhaps derive a canonical extension of a given \( a_n \) and use this to induce an extension of \( F(z) \).

In particular, I think finding an canonical extension of the Liouville \( \lambda \) function onto the complex plane would provide a canonical extension of the Jacobi theta function. This opens a number of deeply fascinating directions of study. For instance, can we study modular forms by studying number theoretic functions extended into the complex plane? I know one implication of this method would be that we can study the primes by looking at extensions of the Moebius function into the complex plane, since that would allow us to study the prime zeta function beyond its natural boundary. There are lots and lots of interesting number-theoretic identities involving Lambert Series, which is basically the form studied in (2). Thus, it conceivable that being able to understand deeply whats going on in (2) could open lots of door for studying number theory. I think a good place to start for studying this would be to look at the Mangolt function. It has a particularly simple relationship \( \sum \Lambda(n) \frac{q^n}{1-q^n} = \sum \ln(n) q^n \). The RHS can be analytically continued using the Lerch Phi function, and this continuation might provide a way to go backwards an induce a value for \( \Lambda \). Probably having a specific case like this to study will be helpful.

Anyway-- thanks for getting all the way down to this point in the post. I'm guessing you probably have lots of questions, and there are probably a few mistakes in my math in the space above-- though I tried to closely check that everything numerically checks out at the least-- so please ask for clarification anywhere that seems confusing or wrong so I can improve my post. Also, any thoughts or insights are appreciated.

Thanks for reading!