# Proving convergence of $\sum_{n=1}^\infty n^{\ln(x)}$

I want to prove that $\sum_{n=1}^\infty n^{\ln(x)}$ is convergent for x in $(0, \frac {1}{e})$ interval and divergent for $x \geq \frac {1}{e}$.

I am lost on how to prove it. Could someone please show me or give me a hint? I assume I would need to use one of the criterias for determining convergence?

$$0<x<\frac{1}{e} \to -\infty<lnx<-1\\lnx=-p\\-\infty<-p<-1\\1<p<+\infty\\\Sigma_{n=1}^{\infty}n^{lnx}=\Sigma_{n=1}^{\infty}n^{-p}=\\\Sigma_{n=1}^{\infty}\frac{1}{n^p} \space,\space 1<p<+\infty$$

Hint:

For $x$ $\in$ $(0,\frac{1}{e})$, $lnx \in (-\infty, -1)$

The integral test says something in particular about series of the form $\sum_{n=1}^\infty \frac{1}{n^p}$ . For $p >1$ these series are convergent. For $p\leq1$ they are strictly divergent.

This is the perfect exercise for the logarithm convergence test: if $x_n > 0 \ \forall n$ and $\lim \dfrac {\ln \frac 1 {x_n}} {\ln n} \left\{ \begin{matrix} >1 \\ =1 \\ <1\end{matrix} \right.$ then $\sum x_n$ is $\left\{ \begin{matrix} \text{convergent} \\ \text{not known} \\ \text{divergent} \end{matrix} \right.$. This is one of the most powerful convergence tests for series with positive terms and I can't understand why it is ignored.

Applying this to your problem, $\lim \dfrac {\ln \frac 1 {n^{\ln x}}} {\ln n} = -\ln x$. According to the above theorem, if $-\ln x > 1$ (i.e. if $x \in (0, \dfrac 1 {\Bbb e} )$) your series converges and if $x > \dfrac 1 {\Bbb e}$ then it diverges.

It remains to study, separately, the case $x = \dfrac 1 {\Bbb e}$ (for which the logarithm test cannot say anything). But this case is really simple, because your series becomes $\sum \dfrac 1 n$, which is known to be divergent.

• In the case of $x = \frac{1}{e}$, the series becomes the Harmonic series. – Aneesh Dec 5 '15 at 15:25
• @moorish: Corrected, thank you, I was in a hurry. – Alex M. Dec 5 '15 at 15:27