Inspiration: IIT-JEE Question

$\displaystyle \lim_{n \rightarrow \infty} \Bigg(\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+2n} \Bigg)$

$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{n} \Bigg(\dfrac{n}{n+1}+\dfrac{n}{n+2}+...+\dfrac{n}{n+2n} \Bigg)$

$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{n} \Bigg(\dfrac{1}{1+\frac{1}{n}}+\dfrac{1}{1+\frac{2}{n}}+...+\dfrac{1}{1+\frac{2n}{n}} \Bigg)$

$\displaystyle \lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{r=1}^{2n} \dfrac{1}{1+\frac{r}{n}} = \int_0^2 \dfrac{1}{1+x} dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle \Bigg(\text{Riemann's sum:} \lim_{n \rightarrow \infty} \dfrac{1}{n} \sum_{r=a}^b f\bigg( \dfrac{r}{n} \bigg) = \int_{\lim_{n \rightarrow \infty} \frac{a}{n}}^{\lim_{n \rightarrow \infty} \frac{b}{n}} f(x) dx \Bigg)$

$=\ln 3 = 1.0986$

The question is basically asking us to evaluate $\displaystyle \large \lim_{n \to \infty} \sum_{k=1}^{2n} \frac{1}{n+k}$

$= \displaystyle \large \lim_{n \to \infty} \sum_{k=1}^{2n} \frac{1}{\left(\frac{1}{2}+\frac{k}{2n}\right)}\cdot \frac{1}{2n} = \lim_{N \to \infty} \sum_{k=1}^{N} \frac{1}{\left(\frac{1}{2}+\frac{k}{N}\right)}\cdot \frac{1}{N}$ , which is a Riemann sum, convertable into an integral.

The Riemann sum is of the form $\displaystyle \large \lim_{n \to \infty} \sum _{k=1}^nf\left(a+kΔx\right)Δx$ , where $Δx = \frac{b-a}{n}$ such that it advances to $\displaystyle \large \int _a^bf\left(x\right)dx$

Comparing this with our Riemann sum, we take $\displaystyle \large f\left(x\right)=\frac{1}{x}$ such that $f\left(a+kΔx\right)Δx = \frac{1}{a+kΔx} \cdot \frac{1}{N}$ and $a+kΔx = \frac{1}{2}+k\left(\frac{1}{N}\right)$ and $Δx = \frac{b-a}{N} = \frac{1}{N}$

This implies that $a=\frac{1}{2}$ and $b=a+1=\frac{3}{2}$

Subsequently, the integral becomes $\int _{\frac{1}{2}}^{\frac{3}{2}}\frac{1}{x}dx = \ln 3 \approx 1.0986$