# Thread: Convergence test for two series

1. ## Convergence test for two series

Hello,

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

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

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

Any help would be appreciated.

2. Originally Posted by thomasdotnet
Hello,

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

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

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

Any help would be appreciated.
Compare the first with

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

and the second with

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$.

3. For the second one, you can also write $\displaystyle log(1 + \frac{1}{n^2}) = \int_1^{1 + \frac{1}{n^2} } \frac{du}{u}$.

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

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

What can you say about the convergence/divergence of the series now?

4. 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 $\displaystyle \frac{1}{n^2}$).

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

5. 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 $\displaystyle \frac{1}{n^2}$).

One thing I cannot completely understand is how you got to $\displaystyle \frac{1}{n\sqrt{n}}$ for the first series.
Consider $\displaystyle \frac{n+1}{\sqrt{n}+1}\sin \frac{1}{n^2}$

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

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

$\displaystyle \sqrt{n} \frac{1}{n^2} = \frac{1}{n \sqrt{n}}$.

6. OK, that makes it clearer. Thank you!