1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.