Trying to show that $\sum_{1}^{\infty} \frac{n^n}{n!} $ diverges.

Question

I have been trying to prove this using the ratio test $|\frac{A_{n+1}}{A_n}|$ , which leads me to this expression: $$\left|\frac{(n+1)^{n+1}}{(n+1)!}\cdot \frac{n!}{n^n}\right|=\left|\frac{(n+1)^{n+1}}{n^n\cdot(n+1)}\right|\cong as \space n\to\infty \frac{n^{n+1}}{n^{n+1}}$$ While it is true that the series diverges, I am getting the wrong ratio value.

Answer 1

Your last step is incorrect. In fact, $$ \frac{(n+1)^{n+1}}{n^n(n+1)} = \frac{(n+1)^n}{n^n} = \left(\frac{n+1}{n}\right)^n = \left(1+\frac1n\right)^n $$ which you might recognize.

In your last step, on the bottom you replaced $n+1$ with $n$, which is okay here because their ratio is $\frac{n+1}{n} = 1+\frac1n \to 1$; but on the top you replaced $(n+1)^{n+1}$ with $n^n$, which is not okay, because their ratio doesn't tend to $1$. Intuitively, the issue is that the exponentiation to the power $n$ makes small differences become large, and the often negligible difference between $n+1$ and $n$ becomes not negligible in such a context.

Answer 2

Hint: Show by induction or inspection that $n^n \geq n!$. Then your terms do not converge to zero so the series necessarily diverges.

Answer 3

$$\frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n}=\frac{(n+1)(n+1)^n}{(n+1)n!}\frac{n!}{n^n}=\frac{(n+1)^n}{n^n}=(1+\frac{1}{n})^n\rightarrow e>1$$