# Convergence test for two series

• September 8th 2009, 12:54 PM
thomasdotnet
Convergence test for two series
Hello,

I am having trouble finding out if the following two series converge.

a) $\sum_{n=1}^{+\infty}{\frac{n+1}{\sqrt{n}+1}\sin{\f rac{1}{n^2}}}$

b) $\sum_{n=1}^{+\infty}{\log\left({1+\frac{1}{n^2}}\r ight)}$

Any help would be appreciated.
• September 8th 2009, 01:35 PM
Jester
Quote:

Originally Posted by thomasdotnet
Hello,

I am having trouble finding out if the following two series converge.

a) $\sum_{n=1}^{+\infty}{\frac{n+1}{\sqrt{n}+1}\sin{\f rac{1}{n^2}}}$

b) $\sum_{n=1}^{+\infty}{\log\left({1+\frac{1}{n^2}}\r ight)}$

Any help would be appreciated.

Compare the first with

$
\sum_{n=1}^{\infty} \frac{1}{n \sqrt{n}}
$

and the second with

$
\sum_{n=1}^{\infty} \frac{1}{n^2}
$
.
• September 8th 2009, 03:29 PM
JG89
For the second one, you can also write $log(1 + \frac{1}{n^2}) = \int_1^{1 + \frac{1}{n^2} } \frac{du}{u}$.

By the MVT of integral calculus we have $\int_1^{1 + \frac{1}{n^2} } \frac{du}{u} = \frac{1}{n^2} \frac{1}{\xi}$, where $1 \le \xi \le 1 + \frac{1}{n^2} \Leftrightarrow \frac{1}{\xi} \le 1$, and so:

$log(1 + \frac{1}{n^2}) = \frac{1}{n^2} \frac{1}{\xi} \le \frac{1}{n^2}$.

What can you say about the convergence/divergence of the series now?
• September 9th 2009, 09:19 AM
thomasdotnet
Thank you very much guys for your help! The second one clearly converges (the integral explanation helped me a lot in understanding why I need to compare this to $\frac{1}{n^2}$).

One thing I cannot completely understand is how you got to $\frac{1}{n\sqrt{n}}$ for the first series.
• September 9th 2009, 10:30 AM
Jester
Quote:

Originally Posted by thomasdotnet
Thank you very much guys for your help! The second one clearly converges (the integral explanation helped me a lot in understanding why I need to compare this to $\frac{1}{n^2}$).

One thing I cannot completely understand is how you got to $\frac{1}{n\sqrt{n}}$ for the first series.

Consider $
\frac{n+1}{\sqrt{n}+1}\sin \frac{1}{n^2}
$

Since $\lim_{n \to \infty} \frac{\sin \frac{1}{n^2}}{\frac{1}{n^2}} = 1\;\;\;$

and for n large $
\frac{n+1}{\sqrt{n}+1} \approx \sqrt{n}$
then we compare with

$
\sqrt{n} \frac{1}{n^2} = \frac{1}{n \sqrt{n}}$
.
• September 9th 2009, 01:11 PM
thomasdotnet
OK, that makes it clearer. Thank you!