Limit of the sequence $\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$ As in the title, we have to calculate the limit $$\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$$ when $n\rightarrow+\infty$. It is easy to see that $$\lim_{n\to\infty} \frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}=0$$ and $$\lim_{n\to\infty} \frac{n}{n^2+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n}=1$$
I don't know how to calculate the limit. I am  pretty sure that we somehow apply the sandwich rule, but I am afraid that there is something similary with $$\lim_{n\to\infty} \frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}=\frac{\pi^2}{6}$$    
 A: Replace all the denominators by $n^2+n$, and then by $n^2+1$, to use squeeze theorem and obtain the answer to be $\frac 12$.
A: Hint:
$$\lim_{n\to\infty} \frac{1}{n^2+1}+\frac{2}{n^2+1}+...+\frac{n}{n^2+1}=\frac{1}{2}$$ and
$$\lim_{n\to\infty} \frac{1}{n^2+n}+\frac{2}{n^2+n}+...+\frac{n}{n^2+n}=\frac{1}{2}$$
A: Convince yourself that
$$\frac1{n^2+n}+\frac2{n^2+n}+\cdots+\frac n{n^2+n}\le \frac1{n^2+1}+\frac2{n^2+2}+\cdots+\frac n{n^2+n} \le$$
$$\le \frac1{n^2}+\frac2{n^2}+\cdots+\frac n{n^2},$$
and remember that $1+2+\cdots+n=\tfrac {n(n+1)}2$. Can you apply the Sandwich rule now?
A: We can rewrite your sum in terms of harmonic numbers and then apply the well-known asymptotics of the latter.
Here we go
$$s = \frac{1}{1+n^2}+ \frac{2}{2+n^2}+ \text{...}+\frac{n}{n+n^2}$$
$$ = (\frac{1}{1+n^2}-1)+ (\frac{2}{2+n^2}-1)+ \text{...}+(\frac{n}{n+n^2}-1) + n*1$$
$$ = n+ (\frac{-n^2}{1+n^2})+ (\frac{2-n^2+2}{2+n^2})+ \text{...}+(\frac{n-n^2-n}{n+n^2}) $$
$$ = n - n^2\left( (\frac{1}{1+n^2})+ (\frac{1}{2+n^2})+ \text{...}+(\frac{1}{n+n^2}) \right)$$
$$ = n-n^2 \sum_{k=1+n^2}^{n+n^2} 1/k$$
This can be written in terms of the harmonic number $H_n = \sum_{k=1}^n 1/k$ as
$$s = n-n^2(H_{n+n^2} - H_{n^2})$$
Now we can calculate the limit $n \to \infty$.
Since in the limit  $x\to\infty$ we have
$$H_x \sim \log(x)+\gamma$$
we find 
$$s \sim n - n^2(\log(n+n^2)-\log(n^2)) $$
$$s \sim n - n^2(\log(1+1/n)) $$
Using the expansion of the $\log$ this becomes finally
$$s \sim n - n^2(\frac{1}{n} - \frac{1}{2n^2}) = 1/2$$
as expected.
A: Another idea is to identify the sum
$$s = \sum _{k=1}^n \frac{k}{k+n^2}$$
in the limit $n\to \infty$ as a Riemann integral.
In fact we can write
$$s = \frac{1}{n} \sum _{k=1}^n \frac{k}{n \left(\frac{k}{n^2}+1\right)}$$
For $n\to\infty$ we can neglect $\frac{k}{n^2}$ against unity:
$$s = \frac{1}{n} \sum _{k=1}^n \frac{k}{n} $$
which asymptotically goes to
$$s \to \int_0^1 x \, dx = 1/2$$
An obvious generalization of the sum gives immediately
$$s_3 = \sum _{k=1}^n \frac{k^2}{k^2+n^3}\to \int_0^1 x^2 \, dx = 1/3 $$
and more generally
$$s_m = \sum _{k=1}^n \frac{k^{m-1}}{k^{m-1}+n^m}\to \int_0^1 x^{m-1} \, dx = 1/m $$
The latter relation holds even for any real $m \gt 0$.
