Solve $\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)$ without using Riemann sums. 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?

 A: We have
\begin{align}
H_n &= 1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n}\\\\
 \Rightarrow  \quad S_n &= H_{2n} - H_n \\\\
& = (H_{2n} - \ln(2n) - \gamma )- (H_n - \ln n - \gamma) + \ln 2\\\\
& = a_n + b_n + \ln 2\to 0+0+\ln 2 = \ln 2\end{align} as $n \to \infty$.
A: One way that does not use the harmonic series is by using the identity $$\sum^{2n}_{i=1} \frac{(-1)^{i-1}}{i}=\sum^n_{i=1}\frac{1}{n+i},$$ which you can find here, for example, and use the power series representation of $\ln(1+x)$ at $x=1$.
A: 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!
A: Starting from $e^x\ge 1+x$ and thus also $e^{-x}\ge 1-x$ one gets
$$
-\ln(1-x)\ge x \ge \ln(1+x)
$$
Now set $x=\frac1{n+k}$ to get
$$
\ln(n+k)-\ln(n+k-1)\ge\frac1{n+k}\ge\ln(n+k+1)-\ln(n+k)
$$
Adding up for the required expression gives telescoping in sums on the left and right, so
$$
\ln(2)=\ln(2n)-\ln(n)\ge\sum_{k=1}^n\frac1{n+k}\ge\ln(2n+1)-\ln(n+1)=\ln\left(2-\frac1{n+1}\right)
$$
