# Thread: Testing series for convergence or divergence

1. ## Testing series for convergence or divergence

$\sum _{n=1}^{\infty }{\frac {n!}{{e}^{{n}^{2}}}}$

I tried the ratio test, from my figuring, I got that it diverges. However, the back of the book says otherwise

I end up getting
${\frac {n+1}{e}}$

I'm pretty sure that's wrong.

2. The ratio test gives $n+1 \over e^{2n+1}$ actually...

3. Originally Posted by CalculusCrazed
$\sum _{n=1}^{\infty }{\frac {n!}{{e}^{{n}^{2}}}}$

I tried the ratio test, from my figuring, I got that it diverges. However, the back of the book says otherwise

I end up getting
${\frac {n+1}{e}}$

I'm pretty sure that's wrong.
If you're using the ratio test, you need to work out

$\lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|$

$= \lim_{n \to \infty}\left|\frac{\frac{(n + 1)!}{e^{(n + 1)^2}}}{\frac{n!}{e^{n^2}}}\right|$

$= \lim_{n \to \infty}\left|\frac{(n + 1)!e^{n^2}}{n!e^{(n + 1)^2}}\right|$

$= \lim_{n \to \infty}\left|\frac{(n + 1)e^{n^2}}{e^{n^2 + 2n + 1}}\right|$

$= \lim_{n \to \infty}\left|\frac{n + 1}{e^{2n + 1}}\right|$

$= \lim_{n \to \infty}\frac{n +1}{e^{2n + 1}}$ since it is positive term...

$= \lim_{n \to \infty}\frac{1}{2e^{2n + 1}}$ by L'Hospital's Rule

$= 0$.

Therefore the series is convergent.

4. Okay cool, I was just doing
${n}^{2}+1$

Thank you so much!