let : $\forall n \in \mathbb{N}$ : $x_n=\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}$ then : $\displaystyle{\sum_{i=1}^{20}}\dfrac{1}{x_i}=?$ let : $\forall n \in \mathbb{N}$ : $x_n=\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}$
then :
$$\sum_{i=1}^{20}\dfrac{1}{x_i}=?$$

my try :
$$\dfrac{1}{\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{13}}+...$$
But boring . 
 A: Let me try.
$$\frac{1}{x_n} = \frac{1}{\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}}$$
$$= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{\frac{4}{n}}$$
$$= \frac{1}{4}\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2}\right)$$
Then, $$\sum_{i=1}^{20}\frac{1}{x_i} = \frac{1}{4}\left(\sqrt{20^2+21^2}-\sqrt{1^2+0^2}\right) = 7.$$
A: I notice that:
$$\begin{align}\frac{1}{\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}} &= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{(1+(1+1/n)^2)-(1+(1-1/n)^2)}\\
&= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{(1+1/n)^2-(1-1/n)^2}\\
&= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{4/n}\\
&= \frac14\left(n\sqrt{1+(1+1/n)^2}-n\sqrt{1+(1-1/n)^2}\right)\\
&= \frac14\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2}\right)
\end{align}$$
That last expression provides for a telescoping sum:
$$\frac14\left(\left(\sqrt{1^2+2^2} - \sqrt{1^2+0^2}\right) + \left(\sqrt{2^2+3^2} - \sqrt{2^2+1^2}\right) + \cdots + \left(\sqrt{19^2+20^2} -
 \sqrt{19^2+18^2}\right) + \left(\sqrt{20^2+21^2} -
 \sqrt{20^2+19^2}\right)\right)$$
which collapses down to:
$$\frac14\left(\sqrt{20^2+21^2} - \sqrt{1^2+0^2}\right)$$
