# Limit comparison test assistance needed please

1. Jan 2, 2015

### Randall

1. The problem statement, all variables and given/known data
Use the limit comparison test to check for convergence or divergence: Sum from n=1 to infinity of ((2n)^2+5)^-3

2. Relevant equations
let lim n to infinity of An/Bn = c
1) if 0<c<infinity then either both converge or both diverge
2) if c=0 and sum Bn converges, so does sum An
3) if c=infinity and sum Bn diverges, so does sum An

3. The attempt at a solution
see attached - I'm a little confused. It seems that it would be easy to pick a comparison series to "force" the answer that you want....

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2. Jan 2, 2015

### Stephen Tashi

Chosing $b_n = 1 + \frac{1}{\sqrt[3]{n} }$ doesn't let you draw any conclusion because $\sum_{n=1}^\infty b_n$ doesn't converge.

3. Jan 2, 2015

### Randall

Doesn't it converge to 1? 1 + 0 = 1? If not, why not? And also what should I choose instead to be bn? Thanks

4. Jan 2, 2015

### Stephen Tashi

Don't confuse the convergence of the sequence $\lim_{n \rightarrow \infty} b_n = 1$ with the convergence of the series $\sum_{n=1}^\infty b_n$. The part 2 in your statement of the limit comparison tests talks about $b_n$ being summable.

My guess is $b_n = (\frac{1}{n^2})^{-3}$.

Edit: Try $b_n = (n^2)^{-3}$

Last edited: Jan 3, 2015
5. Jan 3, 2015

### epenguin

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Last edited: Jan 3, 2015
6. Jan 3, 2015

### epenguin

If you find it very difficult to prove convergence, hedge your bets and try to test for divergence. I would compare with something like 1/n (which I hope you know is divergent.)

7. Jan 3, 2015

### Staff: Mentor

Thread closed. Please stop posting messy images of your work. Most of what you posted has been crossed out, and much of the rest is difficult to read.