$\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?

$$
\begin{align}
&2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\\[16pt]
&=(\sqrt{n}-\sqrt{n-1})-(\sqrt{n+1}-\sqrt{n})\\[9pt]
&=\frac1{\sqrt{n}+\sqrt{n-1}}-\frac1{\sqrt{n+1}+\sqrt{n}}\\
&=\frac{\sqrt{n+1}-\sqrt{n-1}}{(\sqrt{n}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})}\\
&=\frac2{(\sqrt{n}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}+\sqrt{n-1})}\\
&\le\frac1{4(n-1)^{3/2}}
\end{align}
$$
Thus,
$$
\begin{align}
\sum_{n=1}^\infty\left(2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\right)
&\le(2-\sqrt2)+\sum_{n=2}^\infty\frac1{4(n-1)^{3/2}}
\end{align}
$$
and the series converges by the $p$-test.
