TITLE: Solve $\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)$ without using Riemann sums.
QUESTION [5 upvotes]: The natural way (and I think, the easier) to solve
$$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)$$
involves Riemann sums. Multiplying and dividing by $n$, we get
$$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)=\frac{1}{n}\sum_{i=1}^{n}\frac{n}{n+i}=\frac{1}{n}\sum_{i=1}^nf\left(\frac{i}{n}\right)$$
where $f(x)=\frac{1}{1+x}$. The last expression is a Riemann sum associated to the partition $P_n=\{0, \frac{1}{n},\dots,\frac{n}{n}\}$ of $[0,1]$, so by Darboux Theorem
$$\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^nf\left(\frac{i}{n}\right)=\int_0^1\frac{1}{1+x}dx=\ln2$$
However, I'm curious if there are other techniques that can be used to find that limit, so my question is 

Is it possible to find $$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)$$ without using Riemann sums? If so, how?

REPLY [2 votes]: Here is one approach.  Write the sum of interest as 
$$\begin{align}
\sum_{k=n+1}^{2n}\frac1k&=\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac1{2k}\right)\\\\
&=\sum_{k=1}^n\left(\int_0^1 (x^{2k-2}-x^{2k-1})\,dx\right)\\\\
&=\int_0^1 \sum_{k=1}^n \left(x^{2k-2}-x^{2k-1}\right)\,dx\\\\
&=\int_0^1\frac{1-x^{2n}}{1+x}\,dx
\end{align}$$
Using the Dominated Convergence Theorem, we have
$$\begin{align}
\lim_{n\to \infty}\sum_{k=n+1}^{2n}\frac1k&=\int_0^1\lim_{n\to \infty}\left(\frac{1-x^{2n}}{1+x}\right)\,dx\\\\
&=\int_0^1 \frac{1}{1+x}\,dx\\\\
&=\log(2)
\end{align}$$
and we are done!