# Does $\sum_{n=1}^\infty \left(\frac{1}{n} - 1\right)^n$ and $\sum_{n=1}^\infty \left(\frac{1}{n} - 1\right)^{n^2}$ converge?

Determin whether $\sum_{n=1}^\infty \left(\frac{1}{n} - 1\right)^n$ and $\sum_{n=1}^\infty \left(\frac{1}{n} - 1\right)^{n^2}$ converge or diverge.

My thoughts:

For large $n$ one should expect $n^{-1} - 1 \to -1 \neq 0$ raising that to $n$th power should yeald $\{-1,1\}$, again for big $n$, but I'm probably way off here and need some hints tips or better approach to this.

Let $a_n = ( \frac{1}{n} - 1)^{n^2}$

Then $|a_n| = (1- \frac{1}{n})^{n^2} =\exp(n^2 \ln (1-\frac{1}{n})) \sim e^{-n+\frac{1}{2}}$

$\sum e^{-n+\frac{1}{2}}$ converges so $\sum |a_n|$ converges then $\sum a_n$ also converges.

So the second sum does converge.

For the first sum, as already mentioned by H Potter, the general term does not converge to 0 so the sum does not converge.

• If the general term tends to $0$ does it always mean that the series is convergent? – Kushal Bhuyan May 12 '16 at 10:52
• @KushalBhuyan no, for example the famous harmonic series $\sum \frac{1}{n}$ does not converge even if $\frac{1}{n} \rightarrow 0$ – Dark May 12 '16 at 10:53

Note that $(1-\frac1n)^n \rightarrow e^{-1}$ as $n\rightarrow \infty$. Hence, the terms in the first sum don't tend to 0 (they tend to $\pm e^{-1}$ whether $n$ is even or odd) so the sum cannot converge.

I'm currently thinking about the second sum.

• Yes ofcourse, I totally forgot about $(1\pm\frac{1}{n})^n = e^{\pm 1}$, thanks for refreshing my memory. On the second sum as Dark pointed out, converges. Recall that if $\lim_{n\to \infty}|a_n| = 0$ then it converges (on alternating series). Try it with the old trick $e^{\ln|a_n|}$ and L'Hôpital's rule! – Thomas May 12 '16 at 12:05