Can't figure out this limit of a sum $\lim_{n\to\infty}\sum_{k=0}^{2n+1}\frac{n}{n^{2}+k}$ so basically i should find the limit in (+infinity) of this sum $\sum_{k=0}^{2n+1}\frac{n}{n^{2}+k}$ 
I tried to simplify this sum in : $\sum_{k=0}^{2n+1}\frac{1}{n+\frac{k}{n}}$
and started from this equation : $0< k< 2n+1$
and stopped here : $\frac{1}{2+n+\frac{1}{n}}< \frac{1}{n+\frac{k}{n}}< \frac{1}{n}$
now i want to "sum" this inequality to have the first sum but i don't know how and thanks for replying .
 A: With all the steps: if $k$ is in the range $[0,2n+1]$, then $\frac{n}{n^2+k}$ is in the range $\left[\frac{n}{(n+1)^2},\frac{n}{n^2}\right]$, so:
$$ \frac{n(2n+1)}{(n+1)^2}\leq \sum_{k=0}^{2n+1}\frac{n}{n^2+k}\leq \frac{n(2n+1)}{n^2} $$
and squeezing applies, since the limit as $n\to \infty$ of both the RHS and the LHS is just $\color{red}{\large 2}$.
A: If you use harmonic numbers $$S_n=\sum_{k=0}^{2n+1}\frac n{n^2+k}=n\sum_{k=0}^{2n+1}\frac 1{n^2+k}=n \left(H_{(n+1)^2}-H_{(n^2-1)}\right)$$ Now, using for large values of $p$ the asymptotics $$H_p=\gamma +\log \left(p\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^3}\right)$$ apply to each term and continue Taylor expansions to get $$H_{(n+1)^2}-H_{(n^2-1)}=\frac{2}{n}-\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)$$ $$S_n=2-\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.
In the same spirit, we could show that $$T_n=\sum_{k=0}^{an+b}\frac n{n^2+k}=a+\frac{b-\frac{a^2}{2}+1}{n}+\frac{a \left(2 a^2-6 b-3\right)}{6
   n^2}+O\left(\frac{1}{n^3}\right)$$
A: According to the integral test,
$$\int_0^{2n+2}\frac n{n^2+x}\ dx\le\sum_{k=0}^{2n+1}\frac n{n^2+k}\le\frac1n+\int_0^{2n+1}\frac n{n^2+x}\ dx$$
which reduces down to
$$n\ln\left(\frac{n^2+2n+2}{n^2}\right)\le\sum_{k=0}^{2n+1}\frac n{n^2+k}\le\frac1n+2n\ln\left(\frac{n+1}n\right)$$
Thus, by the squeeze theorem,
$$\lim_{n\to\infty}\sum_{k=0}^{2n+1}\frac n{n^2+k}=2$$
