# Thread: Some help/input on Limit Comparison Series Ques.

1. ## Some help/input on Limit Comparison Series Ques.

$\sum_{n=3}^{\infty} \frac{2n+8}{[n ln(n)]^2 +4}$

Can anyone give me some input as to what I could compare this to. At first glance, maybe I could do a limit comparison test with
bn= $\frac{2n}{[n ln(n)]^2}$ , which would simplify to $\frac{2}{n [ln(n)]^2}$

Since n is the dominant term, would I compare it with the harmonic series of $\frac{1}{n}$ ??

Any feedback appreciated. Thanks.

2. ## Re: Some help/input on Limit Comparison Series Ques.

Anyone? I have no idea how to break down [n ln(n)] to find a suitable comparison.

3. ## Re: Some help/input on Limit Comparison Series Ques.

ln(n) is larger than 1 for all n> 1 so that " $[n ln(n)]^2+ 4> n^2+ 4$".

4. ## Re: Some help/input on Limit Comparison Series Ques.

Originally Posted by HallsofIvy
ln(n) is larger than 1 for all n> 1 so that " $[n ln(n)]^2+ 4> n^2+ 4$".
If you replace ln(n) by 1, then the series diverges. On the other hand, $\sum\frac{1}{n(\ln(n))^2}$ converges by the integral test.

5. ## Re: Some help/input on Limit Comparison Series Ques.

Thanks for the responses, the problem make much more sense now.

6. ## Re: Some help/input on Limit Comparison Series Ques.

Originally Posted by Beevo
$\sum_{n=3}^{\infty} \frac{2n+8}{[n ln(n)]^2 +4}$
Well I will the you that we do know that $\sum {\frac{1}{{n{{\left( {\ln (n)} \right)}^2}}}}$ does converge.

But I am unsure of the comparison limit.