TITLE: for what values of x the series converges?
QUESTION [0 upvotes]: For what values of $x$ the series summation$x^{log(n)}$ converges.
I've just used the necessary condition for the convergence of series that $x^{log(n)}$ should tends to zero as $n$ tends to infinity.This is possible only if $|x|<1$.
But this does not surely conclude that the original series converges if $|x|<1$

REPLY [1 votes]: For $x > 0$:
$$\begin{array}{rcl}
\displaystyle \sum_{n=1}^{\infty} x^{\log n}
&=& \displaystyle \sum_{n=1}^{\infty} e^{\log x\log n} \\
&=& \displaystyle \sum_{n=1}^{\infty} n^{\log x} \\
&=& \zeta(-\log x)
\end{array}$$
The last equality holds if and only if the series is convergent, which per this question is if and only if $-\log x > 1$, i.e. $x < \dfrac1e$.
Notice that for the series to be defined, we must have $x \ge 0$.

The series converges if and only if $x \in \left[0,\dfrac1e\right)$.
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