How to get the summation of the series $\sum\limits_{n=0}^{k}{a^{n^b}}$

Question

I want to derive a summation formula for the series: $$\sum\limits_{n=0}^{k}{a^{n^b}}$$

where $a$ and $b$ are two integers

If $b = 0$, then it simply be equal to $ak$ and if $b = 1$ it would just be a geometric series but what about $b \geq 2$, how can I deal with that?

Thanks in advance.

Related, but not a special case due to non-transitivity of exponentiation: Sum of the form $r+r^2+r^4+\dots+r^{2^k} = \sum_{i=1}^k r^{2^i}$ — Dave L. Renfro, Apr 15 '22 at 07:31
With Euler–Maclaurin formula:$$\sum _{n=0}^k a^{n^b}\approx \frac{1}{2} \left(1+a^{k^b}+\frac{2 \left(\Gamma \left(\frac{1}{b}\right)-\Gamma \left(\frac{1}{b},-k^b \ln (a)\right)\right) (-\ln (a))^{-1/b}}{b}\right)$$ — Mariusz Iwaniuk, Apr 15 '22 at 11:41
@Mariusz Iwaniuk is there any way to derive the actual sum? (not just an approximation) — AmirWG, Apr 15 '22 at 12:29
@AmirWG. It is impossible today, but it will be possible in the near future. — Mariusz Iwaniuk, Apr 15 '22 at 12:38
@Mariusz Iwaniuk That's nice to hear. Is there like an ongoing research on this topic? — AmirWG, Apr 15 '22 at 12:52

score 2 · Answer 1 · answered Apr 15 '22 at 02:54

2

The only case is $$\sum\limits_{n=0}^{\color{red}{\infty}}{a^{n^\color{red}{2}}}=\frac{1}{2} (1+\vartheta _3(0,a))$$ where appears Jacobi's theta function

answered Apr 15 '22 at 02:54

Claude Leibovici

260,315

How to get the summation of the series $\sum\limits_{n=0}^{k}{a^{n^b}}$

1 Answers1