Some Sums 2

$\begin{aligned}&\dfrac{1}{\sqrt{r(r+1)}(\sqrt{r+1}+\sqrt{r})}\\&=\dfrac{\sqrt{r+1}-\sqrt r}{\sqrt{r(r+1)}}=\dfrac{1}{\sqrt r}-\dfrac{1}{\sqrt{r+1}}\end{aligned}$ Thus given summation is: $\Large\mathfrak{T}=\displaystyle\sum_{r=1}^{99}\left(\dfrac{1}{\sqrt r}-\dfrac{1}{\sqrt{r+1}}\right)$ $\therefore\large \mathfrak{T}=\left( \begin{aligned} & ~~~~~\left(1-\cancel{\color{#D61F06}{\dfrac{1}{\sqrt 2}}}\right)\\&+\left(\cancel{\color{#D61F06}{\dfrac{1}{\sqrt 2}}}-\cancel{\color{#3D99F6}{\dfrac{1}{\sqrt 3}}}\right)\\&+\left(\cancel{\color{#3D99F6}{\dfrac{1}{\sqrt 3}}}-\cancel{\color{teal}{\dfrac{1}{\sqrt 4}}}\right) \\&\cdots\\&\cdots\\&\cdots\\&+\left(\cancel{\color{goldenrod}{\dfrac{1}{\sqrt 99}}}-\dfrac{1}{\sqrt{100}}\right)\end{aligned} \right)$ $\color{#302B94}{\mathbf{A ~Telescopic ~Series}}$ $\Large =1-\dfrac{1}{10}=\Huge\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{\mathbf{0.9}}}}}}$

$\displaystyle \sum_{r=1}^{99} \dfrac{1}{(r+1)\sqrt{r} + r\sqrt{r+1}} = \sum_{r=1}^{99} \dfrac{(r+1)\sqrt{r}-r\sqrt{r+1}}{(r+1)^{2}r - r^{2}(r+1)} = \sum_{r=1}^{99} \dfrac{1}{\sqrt{r}} - \dfrac{1}{\sqrt{r+1}} = 1 - \dfrac{1}{10} = 0.9$