# Radius of convergence of $\sum_{k=1}^\infty x^{\ln k}$

I need to find the radius of convergence of $$\sum_{k=1}^\infty x^{\ln k}$$

I know how to find the radius of convergence of series of the form $$\sum_{k=1}^\infty a_nx^{n}$$ or $$\sum_{k=1}^\infty a_nx^{2n}$$

But this one has slightly different. Can I have a hint?

• Ratio test seems to work. en.wikipedia.org/wiki/Convergent_series – R zu Oct 6 '18 at 4:29
• This series indeed is equal to $\zeta(-\ln x)$. – Szeto Oct 6 '18 at 4:54
• Since this is not a power series, "radius of of convergence" is not applicable. – zhw. Oct 6 '18 at 23:24

Note that this is not a power series, so there is no such thing as a radius of convergence (no reason to expect the convergence domain to be an interval centered around $$0$$).
If $$x\leq 0$$, then the terms of the series are undefined. Indeed, $$\ln k$$ is not an integer, so there isn't a conventional definition for such a power for a non-positive number. Thus the series is only defined for $$x>0$$.
If $$x>0$$, then $$x^{\ln k}=e^{\ln k \ln x}= k^{\ln x}=\bigg(\frac 1 k\bigg )^{\ln \frac 1x}$$. Therefore the series converges will only converge if $$\ln \frac 1 x >1$$, that is, if $$x.
Conclusion: The domain of convergence of the series is $$(0, e^{-1})$$.
Use $$x^{\ln k}=e^{\ln x\ln k}=k^{\ln x}$$ with $$p$$-series. Indeed the series is $$\zeta(\ln\dfrac1x)$$.