Find the sum of first $20$ terms of a sequence Define a sequence $$a_n=\sqrt{1+\left(1-\frac{1}{n}\right)^2}+\sqrt{1+\left(1+\frac{1}{n}\right)^2}$$ for $n \geq 1$. Find $$\sum_{i=1}^{20}\frac{1}{a_i}$$ Some insight on the approach is highly appreciated. What is the general way of solving such problems? Thanks.
 A: Telescoping is a useful tool.  
$$\frac1{a_n} = \frac1{\sqrt{1+(1-1/n)^2}+\sqrt{1+(1+1/n)^2}} = \frac14(\sqrt{2n^2+2n+1}-\sqrt{2n^2-2n+1})$$
with $f(n) = \frac14\sqrt{2n^2+2n+1}$, this gives $\frac1{a_n} = f(n)-f(n-1)$.  Can you do the rest?
A: First you calculate $\frac{1}{a_i}$ and write it a bit differently:
$\frac{1}{a_n} = 
\frac{\sqrt{1+(1-\frac{1}{n})^2}-\sqrt{1+(1+\frac{1}{n})^2}}{(1+(1-\frac{1}{n})^2)-(1+(1+\frac{1}{n})^2)}
=\frac{\sqrt{1+(1-\frac{1}{n})^2}-\sqrt{1+(1+\frac{1}{n})^2}}{-\frac{4}{n}}
=\frac{1}{4} \left ( \sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2} \right )$.
Now, if you look at $\sum_{n=1}^{20} \frac{1}{4} \left ( \sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2} \right )$, you can see that it is a telescopic sum and thus can be simplified (because most of the terms cancel each other out):
$\sum_{n=1}^{20} \frac{1}{4} \left ( \sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2} \right )=\frac{1}{4} (\sqrt{20^2+21^2}-\sqrt{1^2-0^2})= 7$
A: $$\frac{1}{\sqrt{x}+\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{x-y},$$
so by choosing $x=1+\left(1+\frac{1}{n}\right)^2$, $y=1+\left(1-\frac{1}{n}\right)^2$ we have:
$$ \frac{1}{a_n} = \frac{n}{4}\left(\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}\right) $$
or:
$$ \frac{1}{a_n} = \frac{1}{4}\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2}\right) $$
and our sum turns out to be a telescopic one:
$$ \sum_{n=1}^{20}\frac{1}{a_n}=\frac{1}{4}\left(\sqrt{20^2+21^2}-1\right)=\bbox[5px,border:2px solid #C0A000]{\color{red}{7}}$$
