Is $\sum\limits_{n=1}^{\infty}\frac{n!}{n^n}$ convergent?

I can clearly see that $\dfrac{n!}{n^n}\to 0$ when $n\to\infty$. But how do I know if the sum $$\sum_{n=1}^{\infty}\frac{n!}{n^n}$$ is convergent or not? I know this might be basic, but thank you if anyone can help me.

Solution 1:

This is normally done by using one of the series convergence tests. In your example, ratio test appears applicable, which looks at the limit of ratio of subsequent terms in the series: $R=\lim_{n\to\infty} |a_{n+1}/a_n|$. If you can show that $R<1$, then the original series $\sum_{n=1}^{\infty}a_n$ converges absolutely. For your specific series $$ \frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}/\frac{n!}{n^n}=\frac{n^n}{(n+1)^n}. $$ When $n$ is large $(n+1)^n=\exp(n\ln(n+1))=\exp(n(\ln n+1/n+O(n^{-2})))\approx en^n$, from which it can be concluded that $$ \lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\frac{1}{e}<1. $$ Therefore, your series converges absolutely.

Solution 2:

You can also use Stirling's_formula if you already know it to obtain that $$ \frac{n!}{n^n}\sim \sqrt{2\pi n}\mathrm \cdot e^{-n} $$ and then note that series $\sum\limits_{n=1}^\infty n^k \cdot\mathrm e^{-n}$ converge for any finite $k$.

Solution 3:

Hint: $n!/n^n = (n/n) ((n-1)/n) ((n-2)/n) \cdots (2/n) (1/n)$. Each term in the product is less than or equal to one. So the whole thing is at most $2/n^2$ (for $n \geq 2$).