Proof that $\frac{1}{a_1} +\frac{1}{a_2} +...+\frac{1}{a_{20}}$ is an integer Assume that, for $n\ge1$,$$a_n=\sqrt{1+\left(1+\frac{1}{n}\right)^2 } +\sqrt{1+\left(1-\frac{1}{n}\right)^2 } $$
How to prove that
$$\frac{1}{a_1} +\frac{1}{a_2} +...+\frac{1}{a_{20}}$$ is an integer?
 A: $$\begin{eqnarray}
 \frac{1}{a_n} &=& \frac{1}{\sqrt{1+\left(1+\frac{1}{n}\right)^2} + \sqrt{1+\left(1-\frac{1}{n}\right)^2}} \\ &=& \frac{1}{\sqrt{1+\left(1+\frac{1}{n}\right)^2} + \sqrt{1+\left(1-\frac{1}{n}\right)^2}} \cdot \frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2} - \sqrt{1+\left(1-\frac{1}{n}\right)^2}}{\sqrt{1+\left(1+\frac{1}{n}\right)^2} - \sqrt{1+\left(1-\frac{1}{n}\right)^2}} \\ &=& \frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2} - \sqrt{1+\left(1-\frac{1}{n}\right)^2}}{\left( 1+\left(1+\frac{1}{n}\right)^2 \right) - \left(1+\left(1-\frac{1}{n}\right)^2 \right)} \\
  &=& \frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2} - \sqrt{1+\left(1-\frac{1}{n}\right)^2}}{\frac{4}{n}} \\ &=& \frac{1}{4} \sqrt{n^2+\left(n+1\right)^2} - \frac{1}{4}  \sqrt{n^2+\left(n-1\right)^2} \\
  &=& \frac{1}{4} \sqrt{2 \left(n+\frac{1}{2}\right)^2 + \frac{1}{2}} - \frac{1}{4}  \sqrt{2 \left(n-\frac{1}{2}\right)^2 + \frac{1}{2}} = f(n) - f(n-1)
\end{eqnarray}
$$
Now
$$
   \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_{20}} = f(1) - f(0) + f(2) - f(1) + \cdots + f(20) - f(19) = f(20) - f(0)= 7
$$
A: Rationalizing the fractions yields a fairly nice form:
$$\begin{align*}
\frac1{a_n}&=\frac1{\sqrt{1+\left(1+\frac{1}{n}\right)^2}+\sqrt{1+\left(1-\frac{1}{n}\right)^2}}\cdot\frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}}{\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}}\\
&=\frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}}{\left(1+\frac1n\right)^2-\left(1-\frac1n\right)^2}\\
&=\frac{\sqrt{1+\left(1+\frac{1}{n}\right)^2}-\sqrt{1+\left(1-\frac{1}{n}\right)^2}}{4/n}\\
&=\frac14\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2}\right)\;,
\end{align*}$$
since $$n\sqrt{1+\left(1\pm\frac1n\right)^2}=\sqrt{n^2\left(1+\left(1\pm\frac1n\right)^2\right)}=\sqrt{n^2+\left(n\left(1\pm\frac1n\right)\right)^2}\;.$$
Thus,
$$\begin{align*}
\sum_{n=1}^{20}\frac1{a_n}&=\frac14\sum_{n=1}^{20}\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2}\right)\\
&=\frac14\left(\sum_{n=1}^{20}\sqrt{n^2+(n+1)^2}-\sum_{n=1}^{20}\sqrt{n^2+(n-1)^2}\right)\\
&=\frac14\left(\sum_{n=1}^{20}\sqrt{n^2+(n+1)^2}-\sum_{n=0}^{19}\sqrt{(n+1)^2+n^2}\right)\\
&=\frac14\left(\sqrt{20^2+21^2}+\sum_{n=1}^{19}\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n+1)^2}\right)-\sqrt{1^2-0^2}\right)\\
&=\frac14\left(\sqrt{20^2+21^2}-\sqrt1\right)\\
&=\frac14\left(\sqrt{841}-1\right)\\
&=\frac14(29-1)\\
&=7\;.
\end{align*}$$
If you’re unaccustomed to summation notation, start writing out the sum longhand; you’ll find lots of telescoping.
