A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Determine series convergence or divergence:
a) Sum from 0 to inifinity (ne^(n^2))
anonymous
 4 years ago
Determine series convergence or divergence: a) Sum from 0 to inifinity (ne^(n^2))

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=n \times e ^{n ^{2}}\] is this the function ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[limt \frac{n}{e ^{n2}}\] if we sub value of n then we got \[\frac{\infty}{\infty}\] drive The numerator, and denominator \[\lim \frac{1}{2n \times e^{n^{2}}}\] if we sub infinity \[\frac{1}{\infty} = zero\] also its looks like true by the plot http://www.wolframalpha.com/input/?i=n%2F%28e%5En%5E2%29 Wish I'm right ... you are welcome.

cristiann
 4 years ago
Best ResponseYou've already chosen the best response.0Use a comparison test; for example compare the series with 1/n^2 (which is a convergent series) since (((1/(n²)))/((n/(e^{n²}))))=((e^{n²})/(n³))→∞ you get that eventually (ie beginning with a certain index) 1/n^2 dominates your general term; since both are positive and the bigger one converges, it follows that your series is also convergent The above argument (waleed kha) doesn't work: if the general term goes to 0, then then anything may happen.... for example the general term 1/n goes to 0 but the series diverges
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.