for what values of x the series converges?

Question

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 $|x|<1$

Kenny Lau · Answer 1 · 2017-09-12T16:59:32.113

1

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 $x \in \left[0,\dfrac1e\right)$.

edited Sep 12 '17 at 16:59

answered Sep 12 '17 at 15:39

Kenny Lau

25,049

Note that this is not a power series, so "radius of convergence" isn't quite the right term. – zhw. Sep 12 '17 at 16:44
@zhw. what should I write? – Kenny Lau Sep 12 '17 at 16:50
It's fine now, although your first line should just disappear as it's not helping in any way. – zhw. Sep 12 '17 at 16:54
@zhw. expunged. – Kenny Lau Sep 12 '17 at 16:59

for what values of x the series converges?

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