# Thread: [SOLVED] Sum of infinite series

1. ## [SOLVED] Sum of infinite series

Determine whether the given infinite series converges or diverges. If it converges, find its sum.

$\displaystyle \sum_{n=1}^{\infty}\frac{2n}{\sqrt{4n^2+3}}$

Any help is greatly appreciated!

2. Hi

You should try to take $\displaystyle 4n^2$ out of the square root to find the limit of $\displaystyle \frac{2n}{\sqrt{4n^2+3}}$ when $\displaystyle n\to \infty$.

3. Originally Posted by flyingsquirrel
Hi

You should try to take $\displaystyle 4n^2$ out of the square root to find the limit of $\displaystyle \frac{2n}{\sqrt{4n^2+3}}$ when $\displaystyle n\to \infty$.
How do I take $\displaystyle 4n^2$ out of the square root?

4. You can factor $\displaystyle 4n^2$ : $\displaystyle \sqrt{4n^2+3}=\sqrt{4n^2\left(1+\frac{3}{4n^2}\rig ht)} =\sqrt{4n^2}\sqrt{1+\frac{3}{4n^2}}=\ldots$

5. Originally Posted by flyingsquirrel
You can factor $\displaystyle 4n^2$ : $\displaystyle \sqrt{4n^2+3}=\sqrt{4n^2\left(1+\frac{3}{4n^2}\rig ht)} =\sqrt{4n^2}\sqrt{1+\frac{3}{4n^2}}=\ldots$

I see what I need to do now. Thank you!