$\sum_1^n 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} $ converge or not? how to check if this converge? $$\sum_{n=1}^\infty a_n$$
$$a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}$$
what i did is to show that: 
$$a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n}  - 2\sqrt{n+1} = -2({\sqrt{n+1} - \sqrt{n}}) = -2(({\sqrt{n+1} - \sqrt{n}})\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}) = -2\frac{1}{\sqrt{n+1} + \sqrt{n}} > -2\frac{1}{2\sqrt{n+1}} = \frac{-1}{\sqrt{n+1}} = b_n $$ 
and we know that:
$$\sum_{n=1}^\infty b_n$$
doesnt converge cause 
$$\sum_{n=1}^\infty \frac{1}{\sqrt{n}}$$
doesnt converge
so from here my conclusion is that $\sum_{n=1}^\infty a_n$ doesnt converge
but i know the final answer is that it does converge 
so what am i doing wrong?

Hint
It's a telescoping sum. With $a_n := \sqrt n - \sqrt{n+1}$, $a_n - a_{n-1} = 2\sqrt n - \sqrt{n-1} - \sqrt{n+1}$ so
$$\sum_{i=1}^n 2\sqrt i - \sqrt{i-1} -\sqrt{i+1} = 1 + \sqrt n - \sqrt{n+1}$$

NB that your estimate contains an error in the last step, [fixed now]
$$2 \sqrt n - 2 \sqrt{n+1} = 2 ( \sqrt n - \sqrt{n+1}) \ne 2 (\sqrt n + \sqrt {n+1})$$
Your estimate only proves $S_n > 0$, wich doesn't show anything on its own.
