Seriously serrated series

Let $r^{th}$ term be $T_r$ . So,

$T_r = \sqrt{1 + \frac{1}{r^2} + \frac{1}{(r+1)^2}}$

$T_r = \sqrt{\frac{(r)^2(r+1)^2 + (r+1)^2 + (r)^2)}{(r)^2(r+1)^2}}$

$T_r = \frac{\sqrt{r^4 + 2r^3 + 3r^2 + 2r + 1}}{r(r+1)}$

$T_r = \frac{\sqrt{(r^2 + r + 1)^2}}{r(r+1)}$

$T_r = \frac{(r^2 + r + 1)}{r(r+1)}$

$T_r = 1 + \frac{1}{r(r + 1)}$

$\boxed{T_r = 1 + \frac{1}{r} - \frac{1}{r + 1}}$

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So, sum is

$S = T_1 + T_2 + T_3 + \ldots + T_{2018} + T_{2019}$

$S = \bigg(1 + \frac{1}{1} - \frac{1}{2}\bigg) + \bigg(1 + \frac{1}{2} - \frac{1}{3}\bigg) + \bigg(1 + \frac{1}{3} - \frac{1}{4}\bigg) + \ldots + \bigg(1 + \frac{1}{2018} - \frac{1}{2019}\bigg) + \bigg(1 + \frac{1}{2019} - \frac{1}{2020}\bigg)$

$\color{#3D99F6}{\boxed{S = 2020 - \frac{1}{2020}}}$

The sum can be written as:

$\begin{aligned} S & = \sum_{n=1}^{2019} \sqrt{1 + \frac 1{n^2} + \frac 1{(n+1)^2}} \\ & = \sum_{n=1}^{2019} \sqrt{\frac {n^2(n+1)^2+(n+1)^2+n^2}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{2019} \sqrt{\frac {n^2(n^2+n+1+n)+(n^2+n+1+n)+n^2}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{2019} \sqrt{\frac {(n^2+n+1)(n^2+1)+n^3+n+n^2}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{2019} \sqrt{\frac {(n^2+n+1)(n^2+1)+n(n^2+n+1)}{n^2(n+1)^2}} \\ & = \sum_{n=1}^{2019} \sqrt{\frac {(n^2+n+1)^2}{n^2(n+1)^2}} = \sum_{n=1}^{2019} \frac {n^2+n+1}{n(n+1)} = \sum_{n=1}^{2019} \frac {n^2+n+1}{n^2+n} \\ & = \sum_{n=1}^{2019} \left(1 + \frac 1 {n(n+1)} \right) = \sum_{n=1}^{2019} \left(1 + \frac 1n - \frac 1 {n+1} \right) \\ & = 2019 + 1 - \frac 1{2020} = \boxed{2020-\frac 1{2020}} \end{aligned}$