Hmm. My hypothesis that the double-sum, or at least the exp-series, can only provide continuum sums of functions for "double-exponential type" or less seems to be wrong. Indeed, it seems that if an exp-series exists and converges for a function, then its continuum sum does too.
Consider the triple-exponential function \( f(x) = \exp^3(x) = e^{e^{e^x}} \). This can be expressed as an exp-series
\( e^{e^{e^x}} = \sum_{n=0}^{\infty} a_n e^{nx} \)
where \( a_n \) are the coefficients of the Taylor series of \( e^{e^x} \) at \( x = 0 \) (MacLaurin series). Taking the continuum sum gives,
\( \begin{align}\sum_{n=0}^{x-1} e^{e^{e^n}} &= a_0 x + \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1} \left(e^{nx} - 1\right) \\ &= \left(a_0 - \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1}\right) x + \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1} e^{nx}\end{align} \)
As the coefficients of both sums are smaller than those of the original sum (because \( e^{n} - 1 > 0 \) for all \( n > 0 \)), if the original series converges at 0 and at the point \( x \), so does this (see series comparison test). Since we have an expression for \( e^{e^{e^x}} \), we have achieved its continuum sum and the proof (or disproof, insofar as my original hypothesis that such a series could not yield a continuum sum of something faster than a double exponential, but as a proof this proves even more, namely that any convergent exp-series' continuum sum also converges, unlike the case with Taylor series summed via direct application of Faulhaber's formula) is complete.
As exp-series look to be a special case of nested series, this suggests even 2 layers of nesting may be able to represent tetration and continuum-sum it, though there's still no proof for that. The special case of exp-series themselves do not appear useful for doing tetration with Ansus' formula, however, for two reasons: any function constructed with them is \( 2\pi i \)-periodic, yet tetration seems not to be given pretty much every "good" extension there is -- though they may be able to express tetration for the base \( b = e^{e^{1-e}} \) whose regular tetration is periodic with the required period (and so the exp-series can be recovered via the Fourier series), but a single base isn't very useful. And, we can't even represent \( f(x) = x \) as an exp-series, thus we can't even continuum-sum one exp-series to another exp-series so this is not very useful insofar as trying to iteratively apply Ansus' formula to generate tetrationals goes!
Consider the triple-exponential function \( f(x) = \exp^3(x) = e^{e^{e^x}} \). This can be expressed as an exp-series
\( e^{e^{e^x}} = \sum_{n=0}^{\infty} a_n e^{nx} \)
where \( a_n \) are the coefficients of the Taylor series of \( e^{e^x} \) at \( x = 0 \) (MacLaurin series). Taking the continuum sum gives,
\( \begin{align}\sum_{n=0}^{x-1} e^{e^{e^n}} &= a_0 x + \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1} \left(e^{nx} - 1\right) \\ &= \left(a_0 - \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1}\right) x + \sum_{n=1}^{\infty} \frac{a_n}{e^{n} - 1} e^{nx}\end{align} \)
As the coefficients of both sums are smaller than those of the original sum (because \( e^{n} - 1 > 0 \) for all \( n > 0 \)), if the original series converges at 0 and at the point \( x \), so does this (see series comparison test). Since we have an expression for \( e^{e^{e^x}} \), we have achieved its continuum sum and the proof (or disproof, insofar as my original hypothesis that such a series could not yield a continuum sum of something faster than a double exponential, but as a proof this proves even more, namely that any convergent exp-series' continuum sum also converges, unlike the case with Taylor series summed via direct application of Faulhaber's formula) is complete.
As exp-series look to be a special case of nested series, this suggests even 2 layers of nesting may be able to represent tetration and continuum-sum it, though there's still no proof for that. The special case of exp-series themselves do not appear useful for doing tetration with Ansus' formula, however, for two reasons: any function constructed with them is \( 2\pi i \)-periodic, yet tetration seems not to be given pretty much every "good" extension there is -- though they may be able to express tetration for the base \( b = e^{e^{1-e}} \) whose regular tetration is periodic with the required period (and so the exp-series can be recovered via the Fourier series), but a single base isn't very useful. And, we can't even represent \( f(x) = x \) as an exp-series, thus we can't even continuum-sum one exp-series to another exp-series so this is not very useful insofar as trying to iteratively apply Ansus' formula to generate tetrationals goes!
