A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Convergent or divergent series? sigma (n=1 to infinity) sin (npi/2)/n
anonymous
 5 years ago
Convergent or divergent series? sigma (n=1 to infinity) sin (npi/2)/n

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is a convergent series. It's an alternating series. If you expand on a few terms, you see: S = 1  (1/3) + (1/5)  (1/7) + (1/9)  ... So the series has the form \[\sum_{n=0}^{\infty}(1)^na_n=\]\[\sum_{n=0}^{\infty}(1)^n{\frac{1}{2n+1}}\]where all the a[n] are nonnegative. If the limit of the sequence a[n] equals 0 as n approaches infinity, and the sequence an is monotone decreasing, then the series converges by the Alternating Series Test (otherwise known as the Leibniz Criterion for Alternating Series). The sum is, incidentally, \[\frac{\pi}{4}\]Hope this helps.
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.