Calculating the limit of $\lim\limits_{n \to \infty} (\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ...+\frac{1}{2n})$ Hello everyone how can I calculate the limit of:
$\lim\limits_{n \to \infty} (\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ...+\frac{1}{2n})$
 A: Hint:
Factor out $n$ in all denominators:
$$\frac{1}{n+1} + \frac{1}{n+2}  + …+\frac{1}{2n}=\frac1n\biggl(\frac1{1+\frac1n}+\frac1{1+\frac2n}\dots+\frac1{1+\frac nn}\biggr)$$
and recognise a Riemann sum.
A: It is a very classical question on mathstack... please do some searches before asking a question !
As an answer, consider this :
$$
\sum_{n=k+1}^{2k}\frac1n= \frac 1 n \sum_{n=1}^{k} \frac1 {{ 1+ \frac n k} }
$$
which is related to this integral via a Riemann sum :
$$\int_1^2\frac1x\,dx=\ln 2.$$
A: 
I thought it might be instructive to present an approach that avoids appealing to integrals.  Rather, we make use of elementary arithmetic and the Taylor series for $\log(1+x)=\sum_{k=1}^\infty\frac{(-1)^{k-1}x^k}{k}$.  To that end, we proceed.


Note that we can write the sum of interest as
$$\begin{align}
\sum_{k=1}^n\frac{1}{n+k}&=\sum_{k=n+1}^{2n}\frac{1}{k}\\\\
&=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\\\\
&=\sum_{k=1}^{n}\left(\frac{1}{2k-1}+\frac1{2k}\right)-\sum_{k=1}^n\frac1k\\\\
&=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}
\end{align}$$
Then, using the Taylor series for $\log(1+x)$, evaluated at $x=1$, we find
$$\lim_{n\to\infty}\sum_{k=1}^\infty \frac1{n+k}=\log(2)$$
And we are done!
