Evaluating $\lim_{n \to +\infty}\frac{1}{n}\left({\frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n}}\right)$ How do I evaluate
$$\lim_{n \to +\infty}\frac{1}{n}\left({\frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n}}\right)\ ?$$
I'm stuck. I tried using the sandwich theorem but I was getting nowhere.
Since $n \lt n+k \lt 2n$, I think it should be $ \infty$ .
 A: 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 $$\lim_{n\to\infty}\log(2n)-\log n=\log 2.$$
A: 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}
A: Notice, we have $$\lim_{n \to +\infty}\frac{1}{n}\left({\frac{n}{n+1} + \frac{n}{n+2} + \cdots + \frac{n}{2n}}\right)$$
$$=\lim_{n\to +\infty}\sum_{r=1}^{n}\frac{1}{n}\left(\frac{n}{n+r} \right)$$
$$=\lim_{n\to +\infty}\sum_{r=1}^{n}\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}} \right)$$
Let $\displaystyle \frac{r}{n}=x\implies \lim_{n\to \infty}\frac{1}{n}=dx\to 0$
$$ \text{lower limit of }\ x=\lim_{n\to \infty}\frac{r}{n}=\lim_{n\to \infty}\frac{1}{n}=0$$
$$ \text{upper limit of }\ x=\lim_{n\to \infty}\frac{r}{n}=\lim_{n\to \infty}\frac{n}{n}=1$$
 Now, we have $$\lim_{n\to +\infty}\sum_{r=1}^{n}\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}} \right)=\int_{0}^{1}\left(\frac{1}{1+x} \right)dx$$ $$=\left[\ln (1+x)\right]_{0}^{1}$$$$=\ln (2)-\ln(1)=\ln 2$$
A: 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.
A: Since
$\begin{array}\\
\frac1{k}
 &> \int_k^{k+1} \frac{dx}{x}
> \frac1{k+1}\\
\sum_{k=n+1}^{2n}\frac1{k}
& > \sum_{k=n+1}^{2n}\int_k^{k+1} \frac{dx}{x}
> \sum_{k=n+1}^{2n}\frac1{k+1}\\
or\\
\sum_{k=n+1}^{2n}\frac1{k}
& > \int_{n+1}^{2n+1} \frac{dx}{x}
> \sum_{k=n+1}^{2n}\frac1{k}-\frac1{n+1}+\frac1{2n+1}\\
\end{array}
$
Since
$\int_{n+1}^{2n+1} \frac{dx}{x}
=\ln(2n+1)-\ln(n+1)
=\ln(2-\frac1{n+1})
=\ln(2)-\ln(1-\frac1{2n+2})
$,
if
$d(n)
=\ln(2)-\sum_{k=n+1}^{2n}\frac1{k}
$,
$d(n)
< \ln(1-\frac1{2n+2})
< 0
$
and
$d(n)
>\ln(1-\frac1{2n+2})-\frac1{n+1}+\frac1{2n+1}
$.
Setting
$k=2n+1$
in
$\frac1{k}
 > \int_k^{k+1} \frac{dx}{x}
> \frac1{k+1}
$
we get
$\frac1{2n+1}
> -\ln(1-\frac1{2n+1})
> \frac1{2n+2}
$.
Therefore
$d(n)
>-\frac1{2n+1}-\frac1{n+1}+\frac1{2n+1}
=-\frac1{n+1}
$.
so
$$-\frac1{n+1}
< \ln(2)-\sum_{k=n+1}^{2n}\frac1{k}
< 0.
$$
