Convergence test for two series

**thomasdotnet** · September 8th 2009, 12:54 PM

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.

**Danny** · September 8th 2009, 01:35 PM

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} $ .

**JG89** · September 8th 2009, 03:29 PM

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?

**thomasdotnet** · September 9th 2009, 09:19 AM

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.

**Danny** · September 9th 2009, 10:30 AM

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}}$ .

**thomasdotnet** · September 9th 2009, 01:11 PM

OK, that makes it clearer. Thank you!

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