# limit of the sum $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$ [duplicate]

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)

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

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