Evaluate limit. Let $f : \mathbb R \to \mathbb R$ be differentiable at $x = a$. Evaluate:

$$ \lim_{n\to \infty}\large[{f(a +\frac{1}{n^2})}+{f(a +\frac{2}{n^2})}+...+{f(a +\frac{n}{n^2})}-nf(a)] $$
  Answer: $\   $ $\ \frac{1}{2}f'(a)$

My attempt:
$$ \lim_{n\to \infty}\large[{f(a +\frac{1}{n^2})}-f(a)+{f(a +\frac{2}{n^2})}-f(a)+...+{f(a +\frac{n}{n^2})}- f(a)] $$
I don't know how to proceed from here? Please just give me hint. I want to solve this question by my self. Thank you. 
 A: $[f(a+\frac{1}{n^2})-f(a)]+[f(a+\frac{2}{n^2})-f(a)]+\ldots +[f(a+\frac{n}{n^2})-f(a)]$
$=\dfrac{[f(a+\frac{1}{n^2})-f(a)]}{\frac{1}{n^2}}\times {\frac{1}{n^2}}+\dfrac{[f(a+\frac{2}{n^2})-f(a)]}{\frac{2}{n^2}}\times {\frac{2}{n^2}}+\ldots +\dfrac{[f(a+\frac{n}{n^2})-f(a)]}{\frac{1}{n^2}}\times {\frac{n}{n^2}}$
$=\sum _{i=1}^n [f(a+\frac{k}{n^2})-f(a)]{\frac{k}{n^2}}$
Note that $\lim_{n\to \infty }\dfrac{[f(a+\frac{k}{n^2})-f(a)]}{\frac{k}{n^2}}=f^{'}(a)\to (1)$
ADDED:Following $(1)$;Given $\epsilon>0;\exists m $ such that $n\ge m\implies|\dfrac{[f(a+\frac{k}{n^2})-f(a)]}{{\frac{k}{n^2}}}-f^{'}(a)|<\epsilon\implies $
$(f^{'}(a)-\epsilon)\dfrac{k}{n^2}\le [f(a+\frac{k}{n^2})-f(a)]\le(f^{'}(a)-\epsilon)\dfrac{k}{n^2}\forall k$
Summing over $k$ for all $k=1,2,\ldots n$ we have $(f^{'}(a)-\epsilon)\dfrac{n(n+1)}{2n^2}\le \sum_{k=1}^n[f(a+\frac{k}{n^2})-f(a)]\le(f^{'}(a)-\epsilon)\dfrac{n(n+1)}{2n^2}$
Hence the above expression reduces to $f^{'}(a)[\dfrac{n(n+1)}{2n^2}]\to \dfrac{1}{2}f^{'}(a)$ as $n\to \infty$
@kccu ;Look at the ADDED part to clear your doubts
A: From Taylor's Theorem with the Peano remainder, 

$$\bbox[5px,border:2px solid #C0A000]{f(a+k/n^2)-f(a)=f'(a)\frac k{n^2}+h(k/n^2)\frac k{n^2}} \tag 1$$

where $\displaystyle \lim_{k/n^2\to 0}h(k/n^2)=0$.

Using $(1)$, we can write
$$\begin{align}
\sum_{k=1}^n f(a+k/n^2)-nf(a)&=\frac1{n^2}\sum_{k=1}^n kf'(a)+\frac{1}{n^2}\sum_{k=1}^nkh(k/n^2)\\\\
&=\frac{n(n+1)}{2n^2}f'(a)+\frac1{n^2}\sum_{k=1}^n kh(k/n^2) \tag 2
\end{align}$$
It is easy to see that the limit of the first term on the right-hand side of $(2)$ is $\frac12f'(a)$.  

Next, note that for all $\epsilon>0$ there exists a number $\delta(\epsilon)>0$ such that for $k/n^2<\delta(\epsilon)$, $|h(k/n^2)|\le \epsilon$.  Therefore, given $\epsilon>0$, 
$$\left|\frac{1}{n^2}\sum_{k=1}^n kh(k/n^2)\right|<\epsilon \frac{n(n+1)}{2n^2}\le \epsilon$$  
whenever $k/n^2\le 1/n<\delta$.

Therefore, we have

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty }\sum_{k=1}^n f(a+k/n^2)-nf(a)=\frac12 f'(a)}$$

And we are done!
