<s>[INST] 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.
 [/INST] 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):
$\sum_{n=1}^{20} \frac{1}{4} \left ( \sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2} \right )=\frac{1}{4} (\sqrt{20^2+21^2}-\sqrt{1^2-0^2})= 7$
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