We have the arithmetic sequence in which each term is given by adding a common difference to the previous term. Then there is the geometric sequence in which each term is given by multiplying the previous term by a common ratio. If instead we were to raise an initial term to a common power, then we would have the sequence $$a,a^p,a^{p^2},a^{p^3},\dots$$ $$a_n=a^{p^n}$$ Is there any theory that allows such a sequence to be summed over finitely or infinitely many terms. In other words, is it possible to find a closed form for $$\sum_{k=0}^n a^{p^k}$$ or $$\sum_{k=0}^\infty a^{p^k}$$ for arbitrary $a,p\in\mathbb{C}$ and $n\in\mathbb{N}_0$ in terms of $a,p,n$? Of course, for the infinite sum we need $a,p$ to be chosen such that the sequence converges. Thanks in advance!
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9$\begingroup$ If $p$ is an integer $> 1$, then the resulting sum is lacunary with the natural boundary as the unit circle, due to Ostrowski-Hadamard gap theorem. Since such behavior is hardly expected for elementary functions, I think that such sums do not admit elementary closed-form. Of course, some well-studied special functions are capable of dealing with a special case (such as Jacobi elliptic function). $\endgroup$– Sangchul LeeCommented Jun 29, 2019 at 19:29
1 Answer
For $p > 1$ and $ \Re(x) > 0$ let $$f(x) = \sum_{k=0}^\infty e^{-x p^k}$$ for $\Re(s) > 0$ $$F(s)=\int_0^\infty f(x) x^{s-1}dx = \sum_{k=0}^\infty (p^k)^{-s}\Gamma(s)= \frac{\Gamma(s) }{1-p^{-s}}$$ and with the residue theorem $$f(x) = \frac{1}{2i\pi} \int_{c-i\infty}^{c+i\infty} F(s) x^{-s}ds = \sum Res(F(s) x^{-s}) \\=\sum_{m =1}^\infty \frac{\Gamma(im) x^{-im}+\Gamma(-im)x^{im}}{\log p}+ C+ \sum_{n=1}^\infty \frac{(-1)^n}{n! (1-p^n)} x^n$$
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1$\begingroup$ Sorry, but how is the latter form any better than the original summation? $\endgroup$ Commented Jun 29, 2019 at 21:37
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1$\begingroup$ It is not, it tells different informations. $\Gamma(s)/(1-p^{-s})$ is a much simpler expression than a random special function $\endgroup$– reunsCommented Jun 29, 2019 at 21:55