Evaluating $\lim_{n \to +\infty}\frac{1}{n}\left({\frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n}}\right)$

It can be rewritten as

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

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

And here you recognize a Riemann sum. So the limit is

$$\int_0^1 \frac{1}{1+x} dx$$

Edit : when something can be written as $\frac{1}{n} \sum \cdots$, a sum of Riemann may be hidden inside.

More directly, multiplying by $1/n$ each term inside the parenthesis we get

$$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=$$ $$\lim_{n\to\infty}1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}-1-\frac{1}{2}-\frac{1}{3}-\cdots-\frac{1}{n}=\\\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k},$$ which, since the harmonic series grows logarithmically, is the same as $$\lim_{n\to\infty}\log(2n)-\log n=\log 2.$$

using digamma function we have $$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}=\psi(n+1)-\psi(2n+1)$$ but then using the following asymptotic expansion $$\psi(x)= \ln(x) - \sum_{k=1}^\infty \frac{B_k}{k x^k}$$ with type two Bernoulli numbers $B_k$, we have \begin{align} \lim_{n \to \infty}\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}&=\lim_{n \to \infty}\Big(\psi(n+1)-\psi(2n+1)\Big)\\ &=\ln2 \end{align}