$ \lim \limits_{n \to \infty}(\dfrac{1}{n+1}+\dfrac{1}{n+2}.....\dfrac{1}{n+n})$

Question

How do I find this limit $ \lim \limits_{n \to \infty}(\dfrac{1}{n+1}+\dfrac{1}{n+2}.....\dfrac{1}{n+n})$

$(A)=\log_e1$

$(B)=\log_e2$

$(C)=\log_e3$

$(D)=\log_e4$

Please give me hint or something. What i could see is $ \lim \limits_{n \to \infty}\dfrac{1}{n} \rightarrow 0 $ that is not even in the options.

Answer 1

$\dfrac{1}{n+1}+\dfrac{1}{n+2}.....\dfrac{1}{n+n}= \frac{1}{n}(\frac{1}{1+1/n}+\frac{1}{1+2/n}+....+\frac{1}{1+n/n}) \to \int_0^1 \frac{1}{1+x} dx= \log_e 2$

Answer 2

Your reasoning is invalid, since the number of terms in the limit increases without bound. Regarding a solution, recall that $$\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{n} \approx \log_\mathrm{e} n+\gamma , \quad n \to +\infty,$$ where $\gamma$ denotes the Euler-Mascheroni constant.