Summation vs Radicals

We can use the identity $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$ :

$\sum_{k=1}^{20} \frac1{\sqrt{1+\left(1+\frac1k\right)^2}+\sqrt{1+\left(1-\frac1k\right)^2}}$

$=\sum_{k=1}^{20} \frac{\sqrt{1+\left(1+\frac1k\right)^2}-\sqrt{1+\left(1-\frac1k\right)^2}} {\left(1+\frac1k\right)^2-\left(1-\frac1k\right)^2}$

$=\sum_{k=1}^{20}\frac{k}4 \left(\sqrt{1+\left(1+\frac1k\right)^2}-\sqrt{1+\left(1-\frac1k\right)^2}\right)$

$=\frac14\sum_{k=1}^{20}\left(\sqrt{2k^2+2k+1}-\sqrt{2k^2-2k+1}\right).$

Note that $2(k-1)^2+2(k-1)+1=2k^2-4k+2+2k-2+1=2k^2-2k+1$ . This means the sum telescopes and we find:

$=\frac14\sum_{k=1}^{20}\left(\sqrt{2k^2+2k+1}-\sqrt{2(k-1)^2+2(k-1)+1}\right)$

$=\frac14\left(\sqrt{2(20)^2+2(20)+1}-\sqrt{2(0)^2+2(0)+1}\right)$

$=\frac14(29-1)=\boxed{7}.$

Rationalizing the denominator we have:-
$\displaystyle \sum_{k=1}^{20}\frac{k}4 \left(\sqrt{1+\left(1+\frac1k\right)^2}-\sqrt{1+\left(1-\frac1k\right)^2}\right)\\ \displaystyle =\dfrac 1 4 \sum_{k=1}^{20} \left(\sqrt{k^2+\left(k+1\right)^2}-\sqrt{k^2+\left(k-1\right)^2}\right)\\ Let ~f(k)= \left(\sqrt{k^2+\left(k+1\right)^2}-\sqrt{k^2+\left(k-1\right)^2}\right)\\\therefore~f(n)+f(n+1)\\= \left(\color{#D61F06}{ \sqrt{n^2+\left(n+1\right)^2} }-\sqrt{n^2+\left(n-1\right)^2}~\right)\\+$ $\left( \sqrt{ \left(n+1\right)^2+\left(n+2\right)^2 } - \color{#D61F06}{ \sqrt{\left(n+1\right)^2 +n^2}~}\right)\\\text {So it is a telescopic series. Only the second term of k=1 }\\ \text {and first term of k=20 are not canceled. }\\So ~\displaystyle ~\dfrac 1 4 \sum_{k=1}^{20} f(k)=\dfrac 1 4 *(-1 +\sqrt{20^2+21^2})=~~~~~\color{#3D99F6}{7}$