Closed form solution for partial summation of $\sum_{x=1}^{k} \frac{2^{\frac{1}{x}}}{{x^2}}$

Question

Recently I've been working on solving summations and I found this one to be quite tricky.

$\sum_{x=1}^{k} \frac{2^{\frac{1}{x}}}{{x^2}}$

The integral which this is based off of, can be solved with u substitution is:

$\int_{1}^{k} \frac{2^{\frac{1}{x}}}{{x^2}}dx = -\frac{2^\frac{1}{k} - 2}{\ln(2)}$

How am I able to solve this summation? Is there an analog of u-substitution that I can use for summation, that works similar to how logarithms are used in the integral?

Answer 1

Using Euler-MacLaurin summation

$$\sum_{x=1}^{k} \frac{2^{\frac{1}{x}}}{{x^2}}=C-\frac{1}{k}+\frac{1-\log (2)}{2 k^2}-\frac{1-3\log(2)+\log ^2(2)}{6 k^3}+O\left(\frac{1}{k^4}\right)$$ Adjusting $C$ for $k=100$ gives $C\sim \frac{1309}{468}$.