Does $\displaystyle\sum^{\infty}_{n=1}\left(\frac{n!}{n^n}\right)$ converge or diverge?
I've tried the ratio test, but i'm unsure if I can continue this way.
$$\displaystyle\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\lim_{n\to\infty}\frac{(n+1)n!n^n}{(n+1)^n(n+1)n!}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}$$
If this approach works, how do we proceed? If not, what test might be worth trying?
$$\frac{n^n}{(1+n)^n} = \frac1{\left(\frac{n+1}{n}\right)^n} = \frac{1}{\left(1 + \frac1{n} \right)^n} \longrightarrow \frac1{e}$$
Observe: $\sqrt[n]{\dfrac{n!}{n^n}} = \dfrac{\sqrt[n]{n!}}{n}< \dfrac{1}{n}\cdot\left(\dfrac{1+2+\cdots +n}{n}\right)=\dfrac{n(n+1)}{2n^2}\to \dfrac{1}{2}< 1$. Thus the root test applies and convergence follows.
When you face problems with factorials, it may be good to consider Stirling approximation $$n!\approx \sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$ So, in your case $$u_n=\frac {n!}{n^n}\approx \sqrt{2 \pi } e^{-n} \sqrt{n}$$ $$\frac{u_{n+1}}{u_n}\approx\frac 1e\frac{\sqrt{n+1}}{ \sqrt{n}}\to \frac 1e$$
Let $$a_n=\left(\frac{n!}{n^n}\right)$$ and let $$a_{n+1}=\left(\frac{(n+1)!}{(n+1)^{(n+1)}}\right)$$ By Ratio Test $$\frac{a_{(n+1)}}{a_n}=\left(\frac{(n+1)!}{(n+1)^{(n+1)}}\right).\left(\frac{n^n}{n!}\right)$$ $$\frac{a_{(n+1)}}{a_n}=\frac{(n+1).n!}{(n+1)^n.(n+1)}.\frac{n^n}{n!}$$ $$\frac{a_{(n+1)}}{a_n}=\frac{n^n}{(n+1)^n}$$ $$\frac{a_{(n+1)}}{a_n}=\left(\frac{n}{n+1}\right)^n$$ $$\frac{a_{(n+1)}}{a_n}=\left(\frac{1}{1+\frac{1}{n}}\right)^n$$ $$\frac{a_{(n+1)}}{a_n}=\frac{1}{\left(1+\frac{1}{n}\right)^n}$$ $$\sum_{n=1}^\infty\frac{a_{(n+1)}}{a_n}=\sum_{n=1}^\infty\frac{1}{\left(1+\frac{1}{n}\right)^n}$$ $$\sum_{n=1}^\infty\frac{a_{(n+1)}}{a_n}=\frac{1}{\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)^n}$$ $$\sum_{n=1}^\infty\frac{a_{(n+1)}}{a_n}=\frac{1}{e}$$ So this is convergence by root test.
