Question

Does `a_n=n^n/(n!)` converge?

Answer 1

No, but `a_(n+1)/a_n -> e` as `n->oo`

Explanation:

`a_(n+1) = (n+1)^(n+1)/((n+1)!)`

`=(((n+1)/n)^n n^n (n+1))/(n!(n+1))`

`=((1+1/n)^n n^n)/(n!) = (1+1/n)^n a_n`

Now `lim_(n->oo) (1+1/n)^n = e`

So `lim_(n->oo) (a_(n+1)/a_n) = e`

Answer 2

In addition, the fact that `a_{n+1}/a_{n} -> e > 1` as `n->infty`, as shown already, also implies that the infinite series `sum_{n=1}^{infty}a_{n}=sum_{n=1}^{infty}(n^(n))/(n!)` diverges, by the Ratio Test .

Explanation:

The Ratio Test says that a series `sum_{n=1}^{infty}a_{n}` converges if `lim_{n->infty}|a_{n+1}|/|a_{n}|<1` and diverges if `lim_{n-> infty}|a_{n+1}|/|a_{n}|>1`, assuming these limits exist (or, in the second case, assuming the sequence `|a_{n+1}|/|a_{n}|` diverges to `infty`).