Limit Comparison test

**delgeezee** · March 13th 2012, 12:02 PM

This was on my test today and when we were done me and another guy had a discussion about it. We both did it differently so I was to see who did it correctly.

use limit comparison to find whether the series converges or diverges

$\sum_{k=1}^{\infty}\frac{4}{7+4nln^2n}$

**FernandoRevilla** · March 13th 2012, 04:28 PM

Originally Posted by delgeezee

$\sum_{k=1}^{\infty}\frac{4}{7+4nln^2n}$

I suppose you mean $\sum_{n=2}^{\infty}\frac{4}{7+4n\ln^2n}$ . The series $\sum_{n=2}^{\infty}\frac{1}{n\ln^2n}$ is convergent (use the integral test) and $\lim_{n\to \infty}\left(\frac{4}{7+4n\ln^2n}:\frac{1}{n\ln^2n }\right)=4\neq 0$ , then the given series is convergent.

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