Question

Convergent or divergent? \[\sum_{n = 1}^{\infty} \frac{n!}{e^{n!}}\]

Answer 1

this series converges by Ratio test..

Answer 2

hmmm did you studied taylor series?

Answer 3

yes.. why?

Answer 4

it approaches to zero

Answer 5

\[e^{n!}>>n!\]

Answer 6

that reasoning may not work sometimes. consider:\[\sum_{n = 1}^{\infty} \frac{1}{n}\] 1>>n as n -> infinity but it would still diverge.

Answer 7

Sorry thats 1 << n

Answer 8

The ratio test should be used I think \[\left|\frac{u_{n+1}}{u_n}\right| = \left|\frac{(n+1)!e^{n!}}{n!e^{(n+1)!}}\right| = \left|\frac{(n+1)}{e^{n!n}}\right|\]
since \[e^{n!}/e^{(n+1)!} = e^{n!}/(e^{n!})^{(n+1)} = 1/(e^{n!})^n\]
so \[\lim_{n\rightarrow \infty}\left|\frac{u_{n+1}}{u_n}\right| = 0 < 1\]
so the sum converges.

Answer 9

there is mistake
\[\left| \frac{n+1}{e^{n+1}} \right|\]