<s>[INST] 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$ .
 [/INST] 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
$$-\frac1{n+1}
< \ln(2)-\sum_{k=n+1}^{2n}\frac1{k}
< 0.
$$
</s>