# Infinite Series Assistance

#### kiwilazer

##### New member
Hello, I am a calculus 2 student and finding help is a struggle in this pandemic until I found this forum.

I am really struggling find whether the following series is convergent or divergent. I tried using the limit comparison test but I was not sure if chose the correct variables to find bn. I rewrote the problem at the bottom to be more legible. If there is a different series test you suggest I should try let me know. Thanks in advance.

#### tkhunny

##### Moderator
Staff member
Why did you use sqrt(n) instead of n^2? Try that again. Pick the greatest exponent. That's the one that matters in the limit.

Choosing the right bn is not a fixed idea. You choose what works, First, you may wish to form an opinion. Does your series converge or diverge. If you decide that it might diverge, pick bn from a series that diverges - or not. Just try something. It's just a series and your only trying to determine its convergence properties. You won't break it. The world will not end if you struggle with it or maybe even get it wrong. Go boldly!

#### kiwilazer

##### New member
Okay so I just tried it with bn= n^2/n^(7/2) and it worked to prove by the limit comparison test that the series does diverge because the limit=-12

By LCT the limit must be positive and finite to prove the series converges. Thanks you.

#### HallsofIvy

##### Elite Member
You know, don't you, that $$\displaystyle \frac{n^2}{n^{7/2}}= n^{2- 7/2}= n^{-3/2}$$?

x^a/x^b=x^(a-b)