A community for students.
Here's the question you clicked on:
 0 viewing
Math2400
 one year ago
can someone help me out with these...stating which test for convergence
Math2400
 one year ago
can someone help me out with these...stating which test for convergence

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is based on the answer choices you have typed into the boxes: Number 1 is an alternating series. For all integers n, you either have odd multiples of pi, which make the value of cosine 1, or you have even multiples, which make the value of cosine equal to 1. \(\cos(n\pi), n \in \mathbb{N} = (1)^{n}\) 5 and 6 aren't pseries. You can compare them to pseries, but they themselves are not pseries. 3 has the (1)^n term, so it is an alternating series. Unless you found a way to compare it to a geometric or pseries, which seems like it would be hard to do. 2 and 4 seem fine.

Math2400
 one year ago
Best ResponseYou've already chosen the best response.0So is 5 and 6 Comparison then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, you can compare them to pseries. A couple notes: A pseries will have this form \[a \sum_{n=1}^{\infty}\frac{ 1 }{ n^{p} }\] where a is a constant. Pseries will not have sin, cos, ln, n!, e^n, nothing like that. This is also an example of something which is NOT a pseries \[\sum_{n=1}^{\infty}\frac{ 1 }{ n^{2}+1 }\] Throwing that extra plus 1 in the denominator makes it no longer a pseries. It must only be a constant times a power of n in the denominator. As for dealing with sin and cos, we know they are bounded between 1 and 1. So when you have a sin or a cos, you can always say it is less than or equal to 1. Its a nice comparison to use to help deal with those.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So for number 5, for example, this would be a perfect comparison: \[\frac{ \sin^{2}(5n) }{ n^{2} } \le \frac{ 1 }{ n^{2} }\]

Math2400
 one year ago
Best ResponseYou've already chosen the best response.0hmm for some reason it says one of them is wrong><

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, I see why, didnt catch it. There is a cos(npi) in the denominator, soit would cancel with the (1)^n in the numerator.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So yeah, it actually just all reduces to a pseries. My bad for not noticing.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty}\frac{ (1)^{n}\ln(e^{n} )}{ n^{5}\cos(n\pi) } = \sum_{n=1}^{\infty}\frac{ (1)^{n}n }{n^{5}(1)^{n} } = \sum_{n=1}^{\infty}\frac{ 1 }{ n^{4} }\]
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.