1. ## Limit comparison test

How do you do these using the limit comparison test?

• $\displaystyle \sum_{n=1}^\infty \frac{1}{n(n^2+1)}$

$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$ p=2 > 1 converges

$\displaystyle \lim_{n\rightarrow\infty} \frac {1}{n(n^2+1)} * \frac{n^2}{1} = \lim_{n\rightarrow\infty} \frac{n^2}{n^3+n}$ ??

• $\displaystyle \sum_{n=1}^\infty \frac{3}{n(n+3)}$

$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ p=1 diverges

$\displaystyle \lim_{n\rightarrow\infty}...=\\ \lim_{n\rightarrow\infty} \frac {3n}{n^2+3n}$ ??

Thanks

2. Originally Posted by JJ007
How do you do these using the limit comparison test?

• $\displaystyle \sum_{n=1}^\infty \frac{1}{n(n^2+1)}$
$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$ p=2 > 1 converges

$\displaystyle \lim_{n\rightarrow\infty} \frac {1}{n(n^2+1)} * \frac{n^2}{1} = \lim_{n\rightarrow\infty} \frac{n^2}{n^3+n}$ ?? = ZERO ---> The series converges.

• $\displaystyle \sum_{n=1}^\infty \frac{3}{n(n+3)}$
$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ p=1 diverges

$\displaystyle \lim_{n\rightarrow\infty}...=\\ \lim_{n\rightarrow\infty} \frac {3n}{n^2+3n}$ ?? = ZERO ---> Test failed. Make a limit comparison test with the convergent series: $\displaystyle \sum_{n \geq 1} \frac{1}{n^2}$

Thanks
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