Ratio and root test 
Hi! I am working on some ratio and root test online homework problems for my calc2 class and I am not sure how to completely solve this problem. I guessed on the second part that it converges, but Im not sure how to solve of the value that it converges to. If someone could possibly help me with this problem it would be greatly appreciated. 
 A: You aren't expected to figure out what the value of the series is (although in time you might figure out it has something to do with $e+e^{-1}$). Did you actually compute $\rho$? What is $$\lim_{n\to\infty} \frac{(2n)!}{(2n+2)!}?$$
A: $\rho=\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\rightarrow\infty}|\frac{\frac{1}{(2n+2)!}}{\frac{1}{(2n)!}}|=\lim_{n\rightarrow\infty}|\frac{(2n)!}{(2n+2)!}|=\lim_{n\rightarrow\infty}\frac{1}{(2n+1)(2n+2)}=\lim_{n\rightarrow\infty}\frac{1}{4n^2+5n+2}=0$
I think you may have overlooked the fact that it was a factorial?
A: We have $a_n = \frac{1}{(2n)!}$ and $a_{n+1} = \frac{1}{(2n+2)!}$
\begin{align}\lim_{n \rightarrow \infty} \Big|\frac{a_{n+1}}{a_n}\Big| &= \lim_{n \rightarrow \infty} \Bigg|\frac{(2n)!}{(2n+2)!}\Bigg| \\
&= \lim_{n \rightarrow \infty} \Bigg|\frac{(2n)(2n-1)(2n-2)\cdots}{(2n+2)(2n+1)(2n)(2n-1)(2n-2)\cdots}\Bigg| \\
&= \lim_{n \rightarrow \infty} \Bigg|\frac{1}{(2n+2)(2n+1)}\Bigg| \\
&=0 < 1
\end{align}
Hence, by the Ratio Test, the series $\sum_{n=1}^{\infty} \frac{1}{(2n)!}$ is convergent.
