Limit of a sum of fractions

$a_n=\sum_{k=1}^{n} \frac{1}{n+k}=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}$

How to find $\lim a_n$?

• Look at it as a Riemann sum. – Daniel Fischer Jan 4 '14 at 13:28

With Eulero-Mascheroni : $$\sum_{k = 1}^{n}\frac{1}{k} - \log{n} \rightarrow \gamma$$ $$\sum_{k = 1}^{2n}\frac{1}{k} - \log{2n} \rightarrow \gamma$$ so $$\sum_{k = n+1}^{2n}\frac{1}{k} - \log{2n} +\log{n} \rightarrow 0$$ and then$$\sum_{k = n+1}^{2n}\frac{1}{k} \rightarrow \log{2}$$

HINT

As Daniel Fisher suggested, rewrite 1/(n+k) as (1/n) 1/(1+k/n) and you are done through integration. I am sure you can take from here.

Below is a reformatted version of this answer

HINT

As Daniel Fisher suggested, rewrite $\dfrac 1{n+k}$ as $\dfrac {1}{n} \dfrac {1}{1+\frac k n}$ and you are done through integration. I am sure you can take from here.

• Do you mind if I type your answer in $\LaTeX$ below yours, keeping your version? – Git Gud Jan 4 '14 at 14:08
• @GitGud. I am sorry for that (I even wrote it in my profile). I became almost blind a couple of years ago and I do not "see" what I type. Beside plain ASII, everything is very hard to me and I did not find a way to learn LaTex in my conditions. Be sure I really apologize. I thank you very much if you accept to LaTeXify for me. Cheers. – Claude Leibovici Jan 4 '14 at 14:28
• I read your profile, I knew about your condition before I posted my comment. That's why I'm asking for your permission to add a $\LaTeX$ version of your answer (in your answer), while keeping the original version. – Git Gud Jan 4 '14 at 14:30
• Again, thank you very much for doing it. – Claude Leibovici Jan 4 '14 at 14:31

Brutally speaking, $$\sum_{k=1}^{n}\frac1{n+k}=\left(\sum_{k=1}^{2n}\frac1k\right)-\left(\sum_{k=1}^{n}\frac1k\right)\simeq\ln2n-\ln n=\ln\frac{2n}n=\ln2.$$ Rigourously, we have $$\sum_{k=1}^{n}\frac1{n+k}=\frac1n\cdot\sum_{k=1}^{n}\frac1{1+\frac kn}=\int_0^1\frac{dx}{1+x}=\ln(1+x)|_0^1=\ln2.$$
