Using definite integral to find the limit. I would like someone to verify this exercise for me. Please.
Find the following limit:
$\lim\limits_{n \to \infty}\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n}\right)$
$=\lim\limits_{n \to \infty}\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+2n}\right)$
$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n} \dfrac{1}{n+k}$
$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n} \dfrac{1}{n\left(1+\frac{k}{n}\right)}$
$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n}\left(\dfrac{1}{1+\frac{k}{n}}\cdot\dfrac{1}{n}\right)$
$=\displaystyle\int_{1+0}^{1+2} \frac{1}{x} \,dx$
$=\displaystyle\int_{1}^{3} \frac{1}{x} \,dx$
$=\big[\ln|x|\big] _{1}^3$
$=\ln|3|-\ln|1|$
$=\ln(3)-\ln(1)$
$=\ln(3)$

\begin{align}S_n&=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n}\\
&=\sum_{k=1}^{2n}\frac{1}{n+k}\\
&=\sum_{k=1}^{n}\frac{1}{n+k}+\underbrace{\sum_{k=n+1}^{2n}\frac{1}{n+k}}_{l=k-n}\\
&=\sum_{k=1}^{n}\frac{1}{n+k}+\sum_{k=1}^{n}\frac{1}{2n+l}\\
&=\sum_{k=1}^{n}\left(\frac{1}{n+k}+\frac{1}{2n+k}\right)\\
&=\frac{1}{n}\sum_{k=1}^{n}\left(\frac{1}{1+\frac{k}{n}}+\frac{1}{2+\frac{k}{n}}\right)\\
\end{align}
Therefore,
\begin{align}\lim_{n\rightarrow \infty}S_n&=\int_0^1 \left(\frac{1}{1+x}+\frac{1}{2+x}\right)dx\\
&=\Big[\ln(1+x)+\ln(2+x)\Big]_0^1\\
&=\ln 2+\ln 3-0-\ln 2\\
&=\boxed{\ln 3}
\end{align}
