Limit comparison test

**JJ007** · March 16th 2010, 12:33 PM

How do you do these using the limit comparison test?

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

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

$\lim_{n\rightarrow\infty} \frac {1}{n(n^2+1)} * \frac{n^2}{1} <br /> = \lim_{n\rightarrow\infty} \frac{n^2}{n^3+n}$ ??

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

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

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

Thanks

**General** · March 16th 2010, 12:41 PM

Originally Posted by JJ007

How do you do these using the limit comparison test?

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

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

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

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

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

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

Thanks

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