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?

  • $\begingroup$ Ratio test seems to work. en.wikipedia.org/wiki/Convergent_series $\endgroup$ – R zu Oct 6 '18 at 4:29
  • $\begingroup$ This series indeed is equal to $\zeta(-\ln x)$. $\endgroup$ – Szeto Oct 6 '18 at 4:54
  • $\begingroup$ Since this is not a power series, "radius of of convergence" is not applicable. $\endgroup$ – 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 $0$).

That said, one can still determine the domain of convergence.

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 $x<e^{-1}$.

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 $\zeta(\ln\dfrac1x)$.


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