Representation of $\sum_{n=1}^{\infty}\frac{(k)^{1/n}}{n^k}$ in terms of Riemann zeta function for fixed integer k >1?

Question

I tried many times to write this$\sum_{n=1}^{\infty}\frac{(k)^{1/n}}{n^k}$ in terms of Riemann zeta function but all my attempts are failed, however I conjuctred that the integer part of this sum equal to $k$ or we may can write $\sum_{n=1}^{\infty}\frac{(k)^{1/n}}{n^k}\geq k, k>1$ , Now my question here is : Is it possible to represent that sum in terms of Riemann zeta function ?

Answer 1

For convenience write $b = \ln(k)$. You can write $$ k^{1/n} = \exp(b/n) = \sum_{j=0}^\infty b^j n^{-j}/j!$$ so your sum is $$ \eqalign{\sum_{n=1}^\infty \sum_{j=0}^\infty \frac{b^j} {n^{j+k} j!} &= \sum_{j=0}^\infty \sum_{n=1}^\infty \frac{b^j} {n^{j+k} j!}\cr &= \sum_{j=0}^\infty \frac{b^j}{j!} \zeta(j+k)} $$