# Using definite integral to find the limit.

I would like someone to verify this exercise for me. Please.

Find the following limit:

$$\lim\limits_{n \to \infty}\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n}\right)$$

$$=\lim\limits_{n \to \infty}\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+2n}\right)$$

$$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n} \dfrac{1}{n+k}$$

$$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n} \dfrac{1}{n\left(1+\frac{k}{n}\right)}$$

$$=\lim\limits_{n \to \infty}\sum\limits_{k=1}^{2n}\left(\dfrac{1}{1+\frac{k}{n}}\cdot\dfrac{1}{n}\right)$$

$$=\displaystyle\int_{1+0}^{1+2} \frac{1}{x} \,dx$$

$$=\displaystyle\int_{1}^{3} \frac{1}{x} \,dx$$

$$=\big[\ln|x|\big] _{1}^3$$

$$=\ln|3|-\ln|1|$$

$$=\ln(3)-\ln(1)$$

$$=\ln(3)$$

• Check your bounds of integration again: $k$ ranges from $1$ to $2n$, not $1$ to $n$. – lc2r43 May 14 at 3:44
• @lc2r43 but why? is 3 the upper limit? – gi2302 May 14 at 13:01

Lastly, use $$\;k/n\rightarrow x,\;\;1/n \to dx\;,\;$$ then $$L=\int_{0}^{2} \frac{dx}{1+x}=\ln(1+x)\big|_{0}^{2}=\ln 3\;.$$
Edit: the lower limit is $$x_l=1/n, x_u=2n/n$$, when $$n$$ is large ($$\infty$$} these are 0 and 2
\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}