# [SOLVED] prove that this series to the nth^2 power converges

• Sep 10th 2009, 06:49 PM
hatsoff
[SOLVED] prove that this series to the nth^2 power converges
Quote:

Prove that

$\sum\left(1-\frac{1}{n}\right)^{n^2}$

converges.
My instinct is to use the root test, and try to show that

$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n<1$.

But unfortunately I don't know how to show this either.

Are there any tricks I can use on these nth root proofs which might help here?

Thanks!
• Sep 10th 2009, 07:00 PM
Krizalid
Note that for each $n\ge1$ is $\left( 1-\frac{1}{n} \right)^{n}\le \frac{1}{e},$ and then $\left( 1-\frac{1}{n} \right)^{n^{2}}\le e^{-n},$ besides the series $\sum_{n\ge1}e^{-n},$ converges by the integral test, thus the original series does converge.

----

Of course your idea does work too. In order to prove that $\left( 1-\frac{1}{n} \right)^{n}\to \frac{1}{e}$ as $n\to\infty,$ we have $1-\frac{1}{n}=\frac{n-1}{n}=\frac{1}{\frac{n}{n-1}}=\frac{1}{1+\frac{1}{n-1}}\implies \left( 1-\frac{1}{n} \right)^{n}=\frac{1}{\left( 1+\frac{1}{n-1} \right)^{n}},$

so it's easy to see that $\left( 1+\frac{1}{n-1} \right)^{n}\to e$ as $n\to\infty,$ and hence, the result.
• Sep 10th 2009, 08:04 PM
hatsoff
Thank you very much, but I could still benefit from further assistance.

Once we have $\left( 1-\frac{1}{n} \right)^{n}\le \frac{1}{e}$, the proof is easy for me. However, I am uncertain how you arrived at that inequality.

We have by definition

$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$.

I have no idea how to prove that this series is increasing, but let's say for a moment that I could. In that case, we have, for each $m=n-1$,

$e>\left(1+\frac{1}{m}\right)^m$.

Then algebraically we can show that

$e>\left(\frac{m+1}{m}\right)^m$

flipping:

$\frac{1}{e}<\left(\frac{m}{m+1}\right)^m$.

Substitute back $m=n-1$ to get

$\frac{1}{e}<\left(\frac{n-1}{n}\right)^{n-1}$,

and expand:

$\frac{1}{e}<\left(1-\frac{1}{n}\right)^{n-1}$.

But this isn't good enough to get to our goal, which, again, is:

$\left( 1-\frac{1}{n} \right)^{n}\le \frac{1}{e}$.