If $S_n=\left\{\frac{{\displaystyle 1}}{{\displaystyle 1+\sqrt{n}}}+\frac{{\displaystyle 1}}{{\displaystyle 2+\sqrt{2}n}}+\frac{{\displaystyle 1}}{{\displaystyle 3+\sqrt{3n}}}+\mathrm{.........}+\frac{{\displaystyle 1}}{{\displaystyle n+\sqrt{n^2}}}\right\}$ then $\underset{n\to \infty }{Lt}{S}_n=$

$\begin{array}{l}=\underset{n\to \infty }{lt}\sum _{r=1}^n\frac{1}{r+\sqrt{r.n}}=\underset{n\to \infty }{lt}\frac{1}{n}\sum _{r=1}^n\frac{1}{\frac{r}{n}+\sqrt{\frac{r.n}{n^2}}}\\ =\underset{n\to \infty }{lt}\frac{1}{n}\sum _{r=1}^n\left(\frac{1}{\frac{r}{n}+\sqrt{\frac{r}{n}}}\right)\\ =\underset{0}{\overset{1}{\int }}\frac{1}{x+\sqrt{x}}\\ =\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{x}(\sqrt{x}+1)}dx\\ \sqrt{x}+1=t\\ \frac{1}{2\sqrt{x}}dx=dt\\ \frac{dx}{\sqrt{x}}=2dt\\ =2{\left(\log t\right)}_1^2\\ =2(\log 2-\log 1)\\ =2\log 2\end{array}$