$\lim_{{n \to \infty}} \left( \frac{1}{n} + \frac{1}{\sqrt{n^2 + n}} + \frac{1}{\sqrt{n^2 + 2n}} + \ldots + \frac{1}{\sqrt{n^2 + (n-1)n}} \right)$
$S(n) = \frac{1}{n} + \frac{1}{\sqrt{n^2 + n}} + \frac{1}{\sqrt{n^2 + 2n}} + \ldots + \frac{1}{\sqrt{n^2 + (n-1)n}}$
$S(n) = \int_{0}^{1} \frac{1}{\sqrt{1 + x}} dx$
$\lim_{{n \to \infty}} S(n) = \lim_{{n \to \infty}} \int_{0}^{1} \frac{1}{\sqrt{1 + x}} dx$
Am i on right way?
Yes the result is correct for the details, from the defenetion of Riemann Summation $$ \lim_{n\to\infty} \frac{b-a}{n}\sum_{k=1}^nf \left(a+\frac{b-a}{n}k \right)=\int_a^b f(x) dx$$ denote the formula by N So, $$ N=\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{\sqrt{n^2+(k-1)n}} $$ $$=\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\sqrt{1+\frac{k}{n}-\frac{1}{n}}} $$ $$=\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \left(1+\frac{k}{n}\right)^{-\frac{1}{2}} $$ put a=0 , b=1 in Riemann Summation $$N=\int_0^1\left(1+x\right)^{-\frac{1}{2}} dx=\left(2\left(1+x\right)^{\frac{1}{2}}\right)^1_0 $$ $$ \therefore \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{\sqrt{n^2+(k-1)n}}=2\sqrt{2}-2 $$