# Limit, Riemann Sum, Integration, Natural logarithm

For any natural number $$m$$, $$\lim_{n\rightarrow \infty }\left ( \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots +\frac{1}{mn} \right )=\ln (m)$$.

I tried to prove the statement in the following way.

Proof:

$$\lim_{n\rightarrow \infty }\left ( \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots +\frac{1}{mn} \right )=\lim_{n\rightarrow \infty }\sum_{r=1}^{(m-1)n}\frac{1}{n+r}$$

Dividing the numerator and the denominator of $$\frac{1}{n+r}$$ by $$n$$, we get $$\frac{1/n}{1+r/n}$$.

Therefore,

$$\lim_{n\rightarrow \infty }\sum_{r=1}^{(m-1)n}\frac{1}{n+r}=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r=1}^{(m-1)n}\frac{1}{1+r/n}$$

this is a Riemann sum, so replacing $$\frac{1}{n}$$ with $$dx$$, $$\frac{r}{n}$$ with $$x$$, and integrating between the limits $$x=0$$ and $$x=m-1$$

we get $$$\int_{0}^{m-1}\frac{dx}{1+x}=\ln(1+x)|_{0}^{m-1}=\ln(m)-\ln(1)=\ln(m)\blacksquare$$ Is this a valid way? • Why $$\lim_{n\rightarrow \infty }\left ( \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots +\frac{1}{m+n} \right )=\lim_{n\rightarrow \infty }\sum_{r=1}^{(m-1)n}\frac{1}{n+r}$$ it should be $$=\lim_{n\rightarrow \infty }\sum_{r=1}^{m}\frac{1}{n+r}$$ – gimusi Dec 4 '18 at 8:38 • @gimusi Oops! The denominator of the last term in the expression of the limit should be$mn$instead of$m+n\$. – Hussain-Alqatari Dec 4 '18 at 8:46
• Ah ok! I revise my answer accordingly! – gimusi Dec 4 '18 at 8:47

$$\sum_{r=1}^{(m-1)n}\frac{1}{n+r}=\sum_{r=1}^{mn}\frac{1}{r}-\sum_{r=1}^{n}\frac{1}{r}\sim \ln(mn)-\ln(n)=\ln m$$
A simpler and more direct choice is to write $$\lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{mn} \right) = \lim_{n \to \infty} \sum_{r=n+1}^{mn} \frac{1}{r} = \lim_{n \to \infty} \frac{1}{n} \sum_{r=n+1}^{mn} \frac{1}{r/n} = \int_{x=1}^m \frac{1}{x} \, dx = \log m,$$ but your solution is valid.