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.


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

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


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

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 $


Is this a valid way?

  • $\begingroup$ 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 $$=\lim_{n\rightarrow \infty }\sum_{r=1}^{m}\frac{1}{n+r}$$ $\endgroup$ – user Dec 4 '18 at 8:38
  • $\begingroup$ @gimusi Oops! The denominator of the last term in the expression of the limit should be $mn$ instead of $m+n$. $\endgroup$ – Hussain-Alqatari Dec 4 '18 at 8:46
  • $\begingroup$ Ah ok! I revise my answer accordingly! $\endgroup$ – user Dec 4 '18 at 8:47

Yes your method is correct.

As an alternative by harmonic series

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

  • $\begingroup$ Can you please check again? $\endgroup$ – Hussain-Alqatari Dec 4 '18 at 8:50
  • $\begingroup$ Thanks dear gimusi $\endgroup$ – Hussain-Alqatari Dec 5 '18 at 15:25
  • $\begingroup$ @Hussain-Alqatari You are welcome and you did a great work there! Bye $\endgroup$ – user Dec 5 '18 at 15:27

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.

  • $\begingroup$ Many Thanks heropup! $\endgroup$ – Hussain-Alqatari Dec 5 '18 at 15:22

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