Limit, Riemann Sum, Integration, Natural logarithm

Question

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?

Answer 1

Yes your method is correct.

As an alternative by harmonic series

$$\sum_{r=1}^{(m-1)n}\frac{1}{n+r}=\sum_{r=1}^{mn}\frac{1}{r}-\sum_{r=1}^{n}\frac{1}{r}\sim \ln(mn)-\ln(n)=\ln m$$

Answer 2

A simpler and more direct choice is to write \begin{align}\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,\end{align} but your solution is valid.