For what values of x does the series $\sum_{n=1}^{\infty}\frac{n^x}{x^n} $ converge?

I've attempted to solve this problem but I can't finish up my reasoning - I don't know how to "check" the remaining numbers. Namely:
(1) I showed that this series converges absolutely for $x \in (-\infty,-1) \cup (1, \infty)$
(2) Then, I checked $x=1$ and $x=-1$ - the series does not converge, because $\frac{n^x}{x^n} $ does not approach $0$.
(3) Zero does not work because the series is not defined.
(4). Now, what I am left with is to check the interval $(-1,1)$, but I don't know how to do this. I can cater for $(0,1)$ using the ratio test for the initial series, but what about $(-1,0)$?

  • 3
    $\begingroup$ This series is convergent at $x=-1$ as we have the alternating harmonic series. My intention is use the variable transformation $x\to\frac1x$ and then look at that series. $\endgroup$ – Bumblebee Nov 22 '17 at 22:34

For $x \in (-1, 1), x\neq 0$ we have $x^n \overset{n\to\infty}{\longrightarrow} 0$. So for $x>0$ , we know $\frac{n^x}{x^n}$ does not converge to zero.

For $x<0$ we can write $x=-\frac{1}{y}$ with $y>1$. Then the absolute value of our sequence is

$$\lvert\frac{n^x}{x^n}\rvert = \frac{y^n}{n^\frac1y} \ge \frac{y^n}{n}.$$

To see that this doesn't converge to zero, we can take the $\log$:

$$\log \left( \frac{y^n}{n} \right) = n\log(y) - \log(n)> 0 $$ for $n$ big enough.

That means $\frac{y^n}{n}>1$.

Alternative: (for showing $\frac{y^n}{n}$ doesn't converge to zero)

Recall that $y>1$, so we can write $y=1+ \epsilon$ with $\epsilon > 0$. Now $$\frac{y^n}{n}=\frac{(1+\epsilon)^n}{n}=\frac{\sum_{k=0}^{n}\binom{n}{k}\epsilon^k}{n}=\frac{1+n\epsilon+\sum_{k=2}^{n}\binom{n}{k}\epsilon^k}{n} > \epsilon$$

  • 2
    $\begingroup$ Perhaps not immediate since also $n^x\to 0$ for negative $x$ but nonetheless true. $\endgroup$ – spaceisdarkgreen Nov 22 '17 at 22:57
  • $\begingroup$ @spaceisdarkgreen Yes, absolutely. I edited. Not sure if there is a nicer argument though. $\endgroup$ – blat Nov 22 '17 at 23:33
  • $\begingroup$ I take "exponentials kill powers" to pretty much go without saying (and am embarrassed to not have a slick one-liner for it)... I guess you could apply L'Hospital as many times as necessary... one time in this case since $|x|<1$. $\endgroup$ – spaceisdarkgreen Nov 22 '17 at 23:45

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.