Convergence of series $\sum_{n=1}^\infty a_n$ where $a_n=\frac{(n!)^2 * 3^n}{n^{2n}}$ using ratio test I tried to find the sum of the series $\sum_{n=1}^\infty a_n$, where $$a_n=\frac{(n!)^2 * 3^n}{n^{2n}}$$
and I almost found it with the ratio test. But the problem is at the end.
$$\lim_{n \to \infty} 3 \left(\frac{n}{(n+1)} \right)^{2n} = \frac{3}{e^2} < 1$$
I really don't understand the last step, how do it equal $\frac{3}{e^2}$?
I think I missed some basic stuff here, but can't work it out. If someone can explain or point me in the right direction it would help a lot!
 A: Hint: Note that
$$
\left(\frac{n}{n+1} \right)^{2n}=\frac{1}{\left(\left(1+\frac{1}{n}\right)^n\right)^2}
$$
and that
$$
\left(1+\frac{1}{n}\right)^n\to e\quad\text{as}\quad n\to\infty
$$
A: Recall the formula for e:
$$
e = \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n
$$
A: $$\left(\frac{n}{n+1}\right)^{2n}=\left(\left(\frac{n}{n+1}\right)^n\right)^2$$
$$\left(\frac{n}{n+1}\right)^n=\frac{1}{\left(1+\frac{1}{n}\right)^n}$$
This is a well known limit, and converges to $e$ as $n \to \infty$.
Hence, we have:
$$\frac{3}{\left({\left(1+\frac{1}{n}\right)^n}\right)^2}=\frac{3}{e^2}$$
A: Another way to find the limit.
Considering $$a_n=\frac{(n!)^2 * \alpha^n}{n^{2 n}}\implies \frac{a_{n+1}}{a_n}=\alpha\left(\frac{n}{n+1}\right)^{2 n}$$ $$\log\left( \frac{a_{n+1}}{a_n} \right)=\log(\alpha)+2n\log\left( \frac{n}{n+1} \right)=\log(\alpha)-2n\log\left(1+ \frac{1}{n} \right)\sim \log(\alpha)-2n\times \frac 1n$$ $$\log\left( \frac{a_{n+1}}{a_n} \right)\sim \log(\alpha)-2=\log\left(\frac \alpha {e^2}\right)\implies \frac{a_{n+1}}{a_n}\sim \frac \alpha {e^2}$$
