Series $\sum_{n=1}^\infty \frac{n^n}{(n!)^2}$

Question

I'm trying to study the following series: $$\sum_{n=1}^\infty \frac{n^n}{(n!)^2}$$ All the terms are positive. But I can't solve the limit: $$\lim_{n\rightarrow \infty} \frac{n^n}{(n!)^2}=\lim_{n\rightarrow \infty} \frac{n}{n^2}*\frac{n}{(n-1)^2}*\frac{n}{(n-2)^2}*...*\frac{n}{9}*\frac{n}{4}*n=\lim_{n\rightarrow \infty} \frac{n}{(n-1)^2}*\frac{n}{(n-2)^2}*...*\frac{n}{9}*\frac{n}{4}$$

the first term of the product tends to $\frac{1}{n}$ but I can't say that all the product is $<\frac{1}{n}$ because the other terms are $>\frac{1}{n}$

how can I solve it?

Answer 1

Using Cauchy root test and Stirling's approximation for factorial ($n!\approx (\frac{n}{e})^n$) we have: $$\lim_{n\to\infty}(\frac{n^n}{(n!)^2})^{\frac{1}{n}}=lim_{n\to\infty}\frac{n}{(\frac{n}{e})^2}=lim_{n\to\infty}\frac{e^2}{n}=0$$ which implies on convergence but I don't think the value of it to be calculated at all!

Answer 2

Without sophisticated tools like Stirling's formula, the ratio test works fine here: $$\frac{u_{n+1}}{u_n}=\frac{(n+1)^{n+1}}{\bigl((n+1)!\bigr)^2}\cdot\frac{(n!)^2}{n^n}=\frac{n+1}{(n+1)^2}\biggl(\frac{n+1}n\biggr)^n=\frac1{n+1}\biggl(\frac{n+1}n\biggr)^n $$ which tends to $0$ since the second factor tends to $\mathrm e$.