Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that : $\displaystyle \lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$

the only thing I could think of is that it can be written like this :

$$ \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k+n} =\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\frac{k}{n}+1}=\int_0^1 \frac{1}{x+1} \ \mathrm{d}x=\ln 2$$

is my answer right ? and are there any other method ?(I'm sure there are)

share|improve this question
Looks good to me. –  Fabian Jan 23 '13 at 20:23

2 Answers 2

up vote 8 down vote accepted

$$\int_{k}^{k+1} \frac{1}{x}dx \leq \dfrac{1}{k} \leq \int_{k-1}^{k} \frac{1}{x}dx.$$ $$ \ln\frac{2n+1}{n} \leq \sum_{k=n}^{2n}\frac{1}{k} \leq \ln\frac{2n}{n-1}. $$

share|improve this answer
thanks this is exactly what I was looking for. –  aziiri Jan 23 '13 at 21:02

We are going to use the Euler's constant $$\lim_{n\to\infty}\left(\left(1+\frac{1}{2}+\cdots+\frac{1}{2n}-\ln (2n)\right)-\left(1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n\right)\right)=\lim_{n\to\infty}(\gamma_{2n}-\gamma_{n})=0$$

Hence the limit is $\ln 2$.

share|improve this answer
thank you, but this is way too high for the level of the problem, it was meant for high school and Calculus I student, I'm sure there are other ways. –  aziiri Jan 23 '13 at 20:31
@Chris'ssister do you have some online source for generalization of $\gamma$ to different bases better than wiki? and can you clarify why $\gamma_{2n}=\gamma_{n}$? –  007resu Jan 23 '13 at 21:07
Shouldn't it be Euler-Mascheroni constant? 'cause Euler has a lot of numbers with his name (although if you write "euler constant", google gets you the value of this number). –  JMCF125 Jun 10 '13 at 17:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.