series function

Question

We know that there are some series that can be written in short, for example: $$ \sum_{n=0}^\infty x^n=\frac{1}{1-x},\qquad |x|<1 $$ Is there similar function for $$ \sum_{n=1}^N x^{1/n} $$ or $$ \sum_{n=2}^Nx^{1/n} $$ in terms of $x$ and $N$ ?

Answer 1

Since we are dealing with $x^{1/n}$, I assume that $x\gt0$. In that case, $\lim\limits_{n\to\infty}x^{1/n}=1$, and thus the series $$ \sum_{n=1}^\infty x^{1/n} $$ diverges by the Term Test.

Answer 2

Since $x^{1/n} =e^{\ln x/n} =1+\frac{\ln x}{n} +\frac{\ln^2 x}{2n^2} +... $, $\sum_{n=1}^{M} \left(x^{1/n}-(1+\frac{\ln x}{n})\right) =\sum_{n=1}^{M} x^{1/n}-M-H_M\ln x $ converges, so that could be considered instead.