Does this series converge BY THE (LIMIT) COMPARISON TEST?:
http://www.tiikoni.com/tis/view/?id=0e4b5d4

Question

Does this series converge BY THE (LIMIT) COMPARISON TEST?:
http://www.tiikoni.com/tis/view/?id=0e4b5d4

s3a · Answer

Ok first let me begin for the LIMIT comparison test.

s3a · Answer

If I choose 1/n = b_n then the limit comparison test is inconclusive because the limit = infinity.

s3a · Answer

Am I correct so far?

anonymous · Answer

compare it to $\sum\frac{\sqrt{n}}{n^2}$

anonymous · Answer

since $0\leq \cos^2(n)\leq 1$

s3a · Answer

Oh, wait! I didn't realize that was a sqrt!

anonymous · Answer

i.e. compare it to 
$$\sum\frac{1}{n^{\frac{3}{2}}}$$

s3a · Answer

Ok, so sqrt(n)/n^2 = 1/n^(3/2) which is a converging p series with p = 3/2 > 1. And cos^2 (n) * sqrt(n)/n^2 <= sqrt(n)/n^2 which is converging therefore the initial series converges by the non-limit comparison test! Right?

anonymous · Answer

yes

s3a · Answer

Eeeeexcellent! (Mr. Burns Voice from Simpsons)

s3a · Answer

:P

anonymous · Answer

since $\frac{\cos^2(n)\sqrt{n}}{n^2}\leq \frac{1}{n^{\frac{3}{2}}}$

s3a · Answer

Yes, I see that now, thanks to you! :)

s3a · Answer

I have another 3 mini-problems (1 of which I think I'm having more trouble in than the other two).

Would you mind helping me with those too?

anonymous · Answer

yw

anonymous · Answer

post in a new thread, and i will take a look, although i have to go soon

s3a · Answer

Alright, I'll do it fast.

s3a · Answer

See you soon. :)