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

Nov 5, 2015

No, but ${a}_{n + 1} / {a}_{n} \to e$ as $n \to \infty$

#### 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 ${\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n} = e$

So ${\lim}_{n \to \infty} \left({a}_{n + 1} / {a}_{n}\right) = e$

Nov 5, 2015

In addition, the fact that ${a}_{n + 1} / {a}_{n} \to e > 1$ as $n \to \infty$, 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 ${\lim}_{n \to \infty} | {a}_{n + 1} \frac{|}{|} {a}_{n} | < 1$ and diverges if ${\lim}_{n \to \infty} | {a}_{n + 1} \frac{|}{|} {a}_{n} | > 1$, assuming these limits exist (or, in the second case, assuming the sequence $| {a}_{n + 1} \frac{|}{|} {a}_{n} |$ diverges to $\infty$).