Question

use the integral test to determine whether the infinite series
sum n=1 to infinity of n(e)^-n^2

Answer 1

I've still got some time left, so I can give you a short hint. When integrating, let \(u=n^2\).

Answer 2

and -du=2n

Answer 3

please guys I really need help with this problem its driving me crazy

Answer 4

\[\int_1^\infty \frac{n}{e^{2n}}~dn\]
If the integral converges/diverges, then so does the series.
Letting \(u=n^2~\Rightarrow~du=2n~dn\), so
\[\frac{1}{2}\int_1^\infty e^{-u}~du\]
Does the integral converge or diverge?

Answer 5

\[\bf= \frac{ 1 }{ 2 }\lim_{n \rightarrow \infty}\int\limits_{1}^{n}e^{-u} \ du=\frac{ 1 }{ 2 }\lim_{n \rightarrow \infty}-e^{-n}+e^{-1}\]Can we evalute this limit?

@Brittny

Answer 6

@Brittny ???

Answer 7

im sorry just gimme a sec

Answer 8

im sloving it

Answer 9

it think it will be 1/2 -e^(- infinity)+e-1

Answer 10

it divergent because it goes to infinty , right?

Answer 11

@genius12 is it right ?

Answer 12

@Brittny I think it is best if you look at these two places here:

http://www.youtube.com/watch?v=ojztrQMqLgE

http://cims.nyu.edu/~kiryl/Calculus/Section_8.3--The_Integral_and_Comparison_Tests/The_Integral_and_Comparison_Tests.pdf

They're both extremely helpful.

Answer 13

ok after looking at the problem again. I solved it and got -1/4e^4 which is by the p test converges right ?

Answer 14

ya

Answer 15

@Brittny I suggest you watch the video through the link; the guy does this exact same problem.