<s>[INST] 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 $a_n$. 
 [/INST] 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 $\displaystyle{\sum_n {1 \over n^{3/2}}}$.
</s>