Question

Which test would you run on this series?

Sigma(n=1-infinity) n!/e^n^2

(only n is squared in the denominator, not the entire e^n)

I tried running ratio test but got infinity, which would mean divergence. But the book says the series will converge. Maybe I did ratio test wrongly? I can't figure out what else you would use.

Answer 1

\[\frac{ n! }{ e ^{n ^{2}} }\]

Answer 2

the ratio test works...try it again

Answer 3

\[\frac{ n! }{ e ^{n ^{2}} }\frac{ e ^{(n+1) ^{2}} }{ (n+1)! }\]

Answer 4

if got my fractions slipped

Answer 5

\[\frac{ (n+1)! }{e ^{(n+1)^{2}}}\frac{ e ^{n ^{2}}}{ n! }\]
thats better

Answer 6

\[\frac{ 1 }{ e } \sum_{n=1}^{\infty} n + 1\]

Answer 7

thats what i got. am i simplifying nonsensically? lol

Answer 8

\[(n+1)^2=n^2+2n+1\]

Answer 9

yeaaah. crikey. hang on

Answer 10

\[\lim_{n \rightarrow \infty} \frac{ n }{ e ^{2n+1} } = \lim_{x \rightarrow \infty} \frac{ x }{ e ^{2x+1} }\]

Answer 11

LHR = 1/2 and CV?

Answer 12

wait dang thats probably wrong too

Answer 13

yeah thats not how that works. still working tho

Answer 14

when you use LHR you get 1/2 times the limit of 1 over e^2x+1 which is 1/2 times 0 which does CV according to ratio test. i feel like thats correct but now maybe im doing LHR incorrectly

Answer 15

\[\lim_{n\to\infty}\left|\frac{n+1}{e^{2n+1}}\right|=0<1\]

yes using L'Hospital you get \[\lim_{n\to\infty}\frac{1}{2e^{2n+1}}=0\]

Answer 16

well cool. thanks for walking me through that

Answer 17

np