Question

Determine if \[\sum_{n=1}^{\infty} (-1)^n \left( \frac{n^n}{n!} \right)\] diverges, converges conditionally or converges absolutely.

Answer 1

Use ratio test.

Answer 2

\(|a_{n+1}/a_n|=\frac{(n+1)^{n+1}n!}{(n+1)!n^n}=(\frac{n+1}{n})^n=(1+\frac{1}{n})^n\)
The limit is \(e\). So the series diverges.

Answer 3

Thanks :D

Answer 4

So this is what I did:

\[\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}(n+1)^{n+1}}{(n+1)!}}{\frac{(-1)^{n}n^{n}}{n!}}\right|\]\[\lim_{n\to\infty}\left|\frac{(-1)^{n+1}(n+1)^{n+1}n!}{(-1)^{n}n^{n}(n+1)!}\right|\]\[\lim_{n\to\infty}\left|\frac{(n+1)^{n}}{n^{n}}\right|\]\[\lim_{n\to\infty}\left|\left(\frac{n+1}{n}\right)^{n}\right|\]

How do I continue?

Answer 5

Yes, now \(\frac{n+1}{n}=1+\frac{1}{n}\)

Answer 6

Would you mind giving a detail of how to factor that?

Answer 7

Oh, it should've been obvious :D
\[\frac{n}{n}+\frac{1}{n}\]
So you remove the absolute value just for the fact that \[n\to\infty\]

Answer 8

Since \(n>0\) the expression \(1+\frac{1}{n}\) is positive. So we don't need absolute value.

Answer 9

Then you apply L'hopital correct?

Answer 10

yes you may. But actually that is how we usually define the natural number \(e\).

Answer 11

Ah OK, if you remember the formula. Awesome, thanks again.