Test series for convergance I'm struggling with this series:
$$\sum_{n = 0}^{\infty}\frac{e^n\left ( \left ( n-1! \right ) \right )^2}{\left ( 2n + 1 \right )!}$$
I've tried testing it with a ratio test. As follows:
$$\lim_{n \to \infty} \frac{a_{{n+1}}}{a_{n}}$$
$$\lim_{n \to \infty} \frac{e^{n+1} \cdot \left ( n! \right )^2}{\left ( 2n+3 \right )!}\cdot \frac{\left ( 2n+1 \right )!}{e^n \left ( n-1 \right )!}$$
And in the end, I'm getting:
$$\lim_{n \to \infty} \frac{e^{n+1} \cdot n^2 \cdot \left ( n-1 \right )!}{e^n \left ( 2n+2 \right )\left ( 2n+3 \right )}$$
And this limit is positive infinity so series diverge, but in the Wolfram Alpha I'm getting the opposite result.
No idea if I'm wrong or the wolfram is wrong. So I will be happy for any help.
 A: Let's go back to your initial result for the ratio test, since there's more that can be done there, but with a correction: the $(n-1)!$ term should be squared, and we'll have
$$\lim_{n \to \infty} \frac{e^{n+1} \cdot \left ( n! \right )^2}{\left ( 2n+3 \right )!}\cdot \frac{\left ( 2n+1 \right )!}{e^n \left (( n-1 \right )!)^2}$$
We can do the following:


*

*$e^{n+1}/e^n = e$

*$n!/(n-1)! = n$, and thus their squares simplify to $n^2$

*$(2n+1)!/(2n+3)! = 1/((2n+2)(2n+3))$
Ergo,
$$\lim_{n \to \infty} \frac{e^{n+1} \cdot \left ( n! \right )^2}{\left ( 2n+3 \right )!}\cdot \frac{\left ( 2n+1 \right )!}{e^n \left (( n-1 \right )!)^2} = \lim_{n \to \infty} \frac{e \cdot n^2}{(2n+2)(2n+3)}$$
The denominator simplifies to have $4n^2$ plus a linear and a constant term. In the infinite limit, the $n^2$ dominates and thus
$$\lim_{n \to \infty} \frac{e \cdot n^2}{(2n+2)(2n+3)} = \lim_{n \to \infty}  \frac{e \cdot n^2}{4n^2} = \frac e 4 \approx 0.68$$
Since this value is less than $1$, it gives us convergence by the ratio test, which agrees with Wolfram's result.
