# Testing series for convergence or divergence

• April 27th 2010, 08:09 PM
CalculusCrazed
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.
• April 27th 2010, 08:13 PM
The ratio test gives $n+1 \over e^{2n+1}$ actually...
• April 27th 2010, 08:18 PM
Prove It
Quote:

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.
• April 27th 2010, 08:33 PM
CalculusCrazed
Okay cool, I was just doing
${n}^{2}+1$

Thank you so much! :)