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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)

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marked as duplicate by Jyrki Lahtonen Aug 28 at 7:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4  
Looks good to me. –  Fabian Jan 23 '13 at 20:23

2 Answers 2

up vote 9 down vote accepted

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

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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 $\ln 2$.

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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

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