I came across a weird series with exponential functions and powers of two:
$$\sum_{k=0}^{\infty} \left(1 - e^{-2^{-k}z} \right), z \in \mathbb R_+$$
and have no idea how to solve this (if there even is a closed form expression). WolframAlpha tells me that the series converges (by the ratio test).
My attempts so far have been writing it as $$\lim_{n \to \infty} \left(n + 1 - \sum_{k=0}^n e^{-2^{-k} z}\right)$$ $$ = \lim_{n \to \infty} \left(n + 1 - e^{-z} - e^{-z/2} - e^{-z/4} - e^{-z/8} - ... -e^{-z/2^n}\right)$$ $$ = \lim_{n \to \infty} \left( n + 1 - \left(\left(\frac{1}{e^z}\right) + \left(\frac{1}{e^z} \right)^{1/2} + \left(\frac{1}{e^z} \right)^{1/4} + ... + \left(\frac{1}{e^z} \right)^{1/2^n}\right)\right)$$
which tells me that my question is ultimately about the behavior of the (diverging?) series:
$$ \sum_{k=0}^\infty \left( \frac{1}{e^z} \right)^{1/2^k} = \sum_{k=0}^\infty f(z)^{1/2^k}, f(z) < 1$$.
Any ideas on this one?
Let $f(x)=1-e^{-x}$. Since $f(x)$ is a concave function on $\mathbb{R}^+$, for any $z>0$ $$\sum_{k=0}^{+\infty}f\left(\frac{z}{2^k}\right)\leq f(z/2)=1-e^{-z/2}$$ holds in virtue of Jensen's inequality, hence the series is convergent.
In order to get a "closed form" for the series, just consider the Taylor series of $f(x)$: $$ f(x) = \sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n!}x^n,$$ from which: $$\sum_{k=0}^{+\infty}f\left(\frac{z}{2^k}\right)=\sum_{k=0}^{+\infty}\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n!}\frac{z^n}{2^{nk}}=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n!}\frac{z^n}{2^n-1}.$$