tommy's "linear" summability method

[Caleb]

The link Caleb gave ( https://mathoverflow.net/questions/43869...tical-cont ) me reminded me of this

https://math.stackexchange.com/questions...0-ns-n-s-1

which I posted here too.

The other link he gave :

https://mathoverflow.net/questions/19866...871#198871

There is alot to say about such functions , and I and others have so in the past somewhere.

But the main thing is my method is designed for analytic continuation.

The problem he gave was at a point that was not even continu.

So I see that as a somewhat violation of conditions although I did not clearly stated the conditions.
Without the alternating part that problem probably goes away , even if we hit a natural boundary ...

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By no means is my method considered perfect and I even doubt " perfect summability " exists since we always get undesired properties , divergeance anyway or contradiction results it seems.

It just felt like a nice idea.

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About this 

The other link he gave :

https://mathoverflow.net/questions/19866...871#198871





Giving lim sup or lim inf can be a very tricky thing.

I have at least 2 such questions/conjectures illustrating how tricky it is , posted by my friend mick at mathstackexchange and/or mathoverflow

https://mathoverflow.net/questions/34906...n-13-a-n-2

https://mathoverflow.net/questions/32224...f-m-frac54

https://math.stackexchange.com/questions...-g-frac916

https://math.stackexchange.com/questions...f-m-frac54

and many more and I did not even mention the complicated functions.

Btw the continuum sum has been discussed here.


regards

tommy1729

Actually, that first link (https://math.stackexchange.com/questions...0-ns-n-s-1) is perhaps instructive because I apply a trick similar to your summability method but consideration of the poles actually becomes important. In my answer the pole actually introduces a branch cut. 

But the main thing is my method is designed for analytic continuation.

The problem he gave was at a point that was not even continu.

I'm sorry -- I don't quite understand what you mean here, are you refering to the example I gave of 
\[\sum (-1)^n x^{2^n}\] 
definitely describes a continuous and complex analytic function for \( |x|<1 \). But I do agree that it doesn't satisfy the condition of only having positive terms. I think instead 
\[\sum \frac{(x^{2^n}-1)}{\sqrt{2}^n}\]
should be slightly closer to the type of example you are looking for. We would obtain
\[\sum \frac{(x^{2^n}-1)}{\sqrt{2}^n} = \sum_{k=1}^\infty \frac{\ln(x)^k}{k!2^k(1-2^{k-1/2})}\]
But this doesn't actually match the sum. Instead, we need to pick up the residues. In fact, there is an even a significant residue on the real line that you have to pick up to make the sums even close. For instance, by considering this residue, we get a very good approximation of 
\[\sum \frac{(x^{2^n}-1)}{\sqrt{2}^n} = - \frac{\sqrt{\ln(\frac{1}{x})}}{\Gamma(3/2)}\frac{\pi}{\ln(2)}+\sum_{k=1}^\infty \frac{\ln(x)^k}{k!2^k(1-2^{k-1/2})} \]
The contribution of the extra residues is about of the order \(10^{-7}\), so its barely even perceptable.