Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. 
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n>0\  \forall\  n\ge1$.
  Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.  

I can't show that $a_n > 0\  \forall n\ge1$. I tried using induction but it wouldn't work.
Attempt:
$$
\begin{align}
2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}&=(2\sqrt{n}-(\sqrt{n-1}+\sqrt{n+1}))\cdot {2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\\&={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\end{align}$$
(The calculations are true for sure. No check is desired.)
Denote
$$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\text{ and }b_n={1\over n^{2}}.$$ 
Then 
$$\begin{align*}
\lim_{n\to \infty}{a_n\over b_n} & =\lim_{n\to \infty}n^{2}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=\lim_{n\to \infty}n^{2}\lim_{n\to \infty}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=0
\end{align*}$$
By the comparison test for series convergence, since $\lim_{n\to \infty}{a_n\over b_n}=0$, then if $b_n$ converges, which it does, so does $a_n$. 
 A: $$a_{n}=\left(\sqrt{n}-\sqrt{n-1}\right)-\left(\sqrt{n+1}-\sqrt{n}\right)=\frac{1}{\sqrt{n}+\sqrt{n-1}}-\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
A: I will address both parts of your question: 
Part 1: Proving that $a_n>0$.
From your final form of $a_n$
$$a_n=\frac{2n-2\sqrt{(n-1)(n+1)}}{2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}}$$
we can notice that the denominator is always positive (when $n\ge1$), so it remains to show that $2n>2\sqrt{(n-1)(n+1)}=2\sqrt{n^2-1}$. This is quite easy (thanks to Steven Stadnicki for pointing this out):
$$n^2>n^2-1\implies n>\sqrt{n^2-1}\implies2n>2\sqrt{n^2-1}$$
as desired.
Part 2: Proving that $\sum\limits_{n=1}^\infty a_n$ converges by the limit comparison test.
Unfortunately your solution is not quite correct as there is an error in your limit computation, and you should get $\lim\limits_{n\rightarrow\infty}n^2a_n=\infty$ which doesn't work for us. However, we can instead use $n^{3/2}$, and find that $\lim\limits_{n\rightarrow\infty}n^{3/2}a_n=\frac{1}{4}$, which proves the sum converges because $\displaystyle\sum_{n=1}^\infty\frac{1}{n^{3/2}}$ converges.
Limit Computations:
First we note 
$\begin{align}a_n&=\frac{2n-2\sqrt{(n-1)(n+1)}}{2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}}
\\&=\frac{2\left(n-\sqrt{n^2-1}\right)}{2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}}
\\&=\frac{2\left(n-\sqrt{n^2-1}\right)}{2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}}\times\frac{n+\sqrt{n^2-1}}{n+\sqrt{n^2-1}}
\\&=\frac{2}{\left(2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}\right)\left(n+\sqrt{n^2-1}\right)}
\end{align}$
Hence
$\begin{align}\lim\limits_{n\rightarrow\infty} n^2a_n&=\lim\limits_{n\rightarrow\infty}\frac{2n^2}{\left(2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}\right)\left(n+\sqrt{n^2-1}\right)}
\\&=\lim\limits_{n\rightarrow\infty}\frac{2\sqrt{n}}{\left(2+\sqrt{1-\frac{1}{n}}+\sqrt{1+\frac{1}{n}}\right)\left(1+\sqrt{1-\frac{1}{n^2}}\right)}
\\&=\infty
\end{align}$
while
$\begin{align}\lim\limits_{n\rightarrow\infty} n^{3/2}a_n&=\lim\limits_{n\rightarrow\infty}\frac{2n^{3/2}}{\left(2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}\right)\left(n+\sqrt{n^2-1}\right)}
\\&=\lim\limits_{n\rightarrow\infty}\frac{2}{\left(2+\sqrt{1-\frac{1}{n}}+\sqrt{1+\frac{1}{n}}\right)\left(1+\sqrt{1-\frac{1}{n^2}}\right)}
\\&=\frac{1}{4}
\end{align}$
A: We have, $$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1} =\frac{1}{\sqrt{n} + \sqrt{n-1}} -  \frac{1}{\sqrt{n+1} + \sqrt{n}} \ge 0$$
and $\displaystyle \sum a_n$ telescoping!
A: You don't have to see the formula. The $n$th term of the sequence is of the form $f(n+1) - 2f(n) + f(n-1)$, where $f(n) = -\sqrt{n}$. 
This is a "difference of differences"; $f(n+1) - 2f(n) + f(n-1) = g(n+1) - g(n)$, where $g(n) = f(n) - f(n-1)$. So the infinite sum is telescoping and will equal $\lim_{n \rightarrow \infty} g(n) - g(1)$ once you show the limit exists.

I thought of a less tricky way to show this. Write
$$a_n = 2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1} = \sqrt{n}\bigg(2 - \sqrt{1 - {1 \over n}} - \sqrt{1 + {1 \over n}}\bigg)$$
Use Taylor series on the square roots to get
$$a_n =  \sqrt{n}\bigg[2 - \bigg(1 - {1 \over 2n} + O\bigg({1 \over n^2}\bigg)\bigg) - \bigg(1 + {1 \over 2n} + O\bigg({1 \over n^2}\bigg)\bigg)\bigg]$$
$$= \sqrt{n} O\bigg({1 \over n^2}\bigg)$$
$$= O\bigg({1 \over n^{3/2}}\bigg)$$
Thus the series converges in comparison with $\displaystyle{\sum_n {1 \over n^{3/2}}}$.
A: Copied from my answer to this near-duplicate:
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
$$\sum_{i=1}^n 2\sqrt i - \sqrt{i-1} -\sqrt{i+1} = 1 + \sqrt n - \sqrt{n+1} \stackrel{n\to\infty}\longrightarrow 1$$
This even allows you to obtain the value.
A: Write $$a_n = \frac {4} {(2 \sqrt {n} + \sqrt {n - 1} + \sqrt {n + 1}) (2n + 2 \sqrt {n^2 - 1})} \sim \frac {1} {4 n \sqrt {n}}.$$ Then, $$\sum a_n \sim \frac {1} {4} \zeta (\frac {3} {2}).$$
