I am trying to find radius of convergence of:

$$\sum_{n=1}^{\infty} n^n x^n$$

I tried using logs to get rid of the power as it proved to be a hindrance in the ratio test. Letting

\begin{align} y &= n^n x^n \\ &= e^{n \ln(nx)} \end{align}

By the ratio test, we obtain

$$\frac{e^{(n+1) \ln{((n+1)x)}}}{e^{n \ln{(nx)}}} = e^{(n+1) \ln{((n+1)x)} - n \ln{(nx)}}$$

Now I am stuck. I am thinking of solving for the limit of the last expression as $ n\to\infty$ . But this would seem too complicated to follow. I am sure that there is a better solution to this. Could someone please enlighten me?

  • $\begingroup$ Are you familiar with the root test? $\endgroup$ – carmichael561 Sep 26 '16 at 5:02
  • $\begingroup$ Nope. I was only taught the ratio test. $\endgroup$ – LanceHAOH Sep 26 '16 at 5:03
  • 1
    $\begingroup$ I believe the radius of convergence is $0$. $\endgroup$ – Will Sherwood Sep 26 '16 at 5:03
  • $\begingroup$ Yes. But I have difficulty figuring out why. Even after I compute the limit from the above, I get $ \infty $. I think I went wrong somewhere. $\endgroup$ – LanceHAOH Sep 26 '16 at 5:04
  • $\begingroup$ $$\frac{(n+1)^{n+1}}{n^{n}} = \left ( 1 + \frac{1}{n} \right )^{n} (n + 1) \to \infty, n \to \infty$$ Hence, by the ratio test, the radius of convergence is zero. $\endgroup$ – mattos Sep 26 '16 at 5:10

This is done easily using the Root Test. But since you said you did not know it, we will use the Ratio Test: $$ \lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|=\lim_{n \to \infty} \left|\dfrac{(n+1)^{n+1}x^{n+1}}{n^nx^n}\right|=\lim_{n \to \infty} \left| \dfrac{(n+1)^n(n+1)}{n^n} \cdot \dfrac{x^{n+1}}{x^n}\right| $$ But pull things together under the same exponent: $$ \lim_{n \to \infty} \left| \left(\dfrac{n+1}{n}\right)^n(n+1) \dfrac{x^{n+1}}{x^n} \right|=\lim_{n \to \infty} \left| \left(1+\dfrac{1}{n}\right)^n(n+1) \,x\right|= |x| \lim_{n \to \infty}\left( \left(1+\dfrac{1}{n}\right)^n \, (n+1) \right) $$ But $\displaystyle\lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)^n=e$ and $\displaystyle\lim_{n \to \infty} (n+1)=\infty$. So the above limit will diverge unless $x=0$. Therefore, the radius of convergence is $0$ and the 'interval' of convergence is simply $\{0\}$.

A good problem to try for yourself to see if you can do this yourself is $$ \sum_{n=1}^\infty \dfrac{n^n}{n!} x^n $$ This series has a finite nonzero radius of convergence. You can check your answer here.

  • $\begingroup$ +1 Clear explanation! Thanks for your helpl! I understand now. $\endgroup$ – LanceHAOH Sep 26 '16 at 5:27

We can use the easiest test: If a series converges, then the sequence of terms converges to zero. But $$ |n^n x^n| = |nx|^n > 1 $$ for all $n > 1/|x|$, unless $x = 0$.


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