Evaluate the limit of $\sum\limits _ { k = 0} ^ { n } \mathrm{arctg} \frac { k + 1} { n ^ { 2} }$ when $n\to\infty$ 
Evaluate $$\lim _ { n \rightarrow \infty } \sum _ { k = 0} ^ { n } \mathrm{arctg} \frac { k + 1} { n ^ { 2} }$$

At first I thought this was a Riemann sum, but I couldn't insert the $1/n$ and get the right form. I also tried to write it out, and it definitely looks like it would converge, but I'm not sure how to approach it.
 A: First, recall that 
$$\arctan(x)=x+O(x^3)$$
as $x\to 0$. 
Then, we have
$$\arctan\left(\frac{k+1}{n^2}\right)=\frac{k+1}{n^2}+O\left(\frac{(k+1)^3}{n^6}\right)$$
as $\frac{k+1}{n^2}\le \frac{n+1}{n^2}\to 0$.
Since $\sum_{k=0}^{n}k^3=O(n^4)$, only the linear term contributes in the limit as $n\to \infty$.  Therefore, 
$$\lim_{n\to \infty}\sum_{k=0}^n\arctan\left(\frac{k+1}{n^2}\right)=\lim_{n\to \infty}\sum_{k=0}^n\left(\frac{k+1}{n^2}\right)=\frac12$$
A: Set $\dfrac{1}{n^2} = m$
We get $\arctan\left(\dfrac{k+1}{n^2}\right)=\arctan\left((k+1) m^2\right)$
Using MacLaurin series we have
$\arctan\left((k+1) m^2\right)=(k+1) m^2+O\left(m^6\right)$ 
that is 
$$\arctan\left(\dfrac{k+1}{n^2}\right)\approx \dfrac{k+1}{n^2}$$
which means that 
$$\sum _{k=0}^n \arctan\left(\frac{k+1}{n^2}\right)\approx\sum _{k=0}^n \frac{k+1}{n^2},\text{ as }n\to\infty$$
as 
$$\sum _{k=0}^n (k+1)=\frac{1}{2} (n+1) (n+2)$$
we have
$$\sum _{k=0}^n \frac{k+1}{n^2}=\frac{(n+1) (n+2)}{2 n^2}$$
and
$$\lim_{n\to \infty } \, \frac{(n+1) (n+2)}{2 n^2}=\frac{1}{2}$$
