How to show that $\lim_{n\to\infty}\left\{f(c+\frac{1}{n^2})+f(c+\frac{2}{n^2})+\cdots+f(c+\frac{n}{n^2})-nf(c)\right\}=\frac{1}{2} f'(c)$ If $f:\mathbb{R}\to\mathbb{R}$ be a differentiable at $x=c$ then how can I show that $$\lim_{n\to\infty}\left\{f(c+\frac{1}{n^2})+f(c+\frac{2}{n^2})+\cdots+f(c+\frac{n}{n^2})-nf(c)\right\}=\frac{1}{2} f'(c)$$
I had no clue which result should I use to prove this. Please give some hints. 
Thanks!
 A: Hint: $f(c+h) = f(c) + f'(c)h +hr(h),$ where $\lim_{h\to 0} r(h) = 0.$
A: $\lim_{n\to\infty}\left\{f(c+\frac{1}{n^2})+f(c+\frac{2}{n^2})+\cdots+f(c+\frac{n}{n^2})-nf(c)\right\}=\lim_{n\to\infty}\left\{f(c)+\frac{f'(c)}{n^2}+f(c)+\frac{2f'(c)}{n^2}+...+f(c)+\frac{nf'(c)}{n^2}-nf(c)\right\}=
\lim_{n\to\infty}\left\{nf(c)+f'(c)\times\frac{1+2+...+n}{n^2}-nf(c)\right\}=
\lim_{n\to\infty}\left\{f'(c)\times\frac{n(n+1)}{2n^2}\right\}=
\frac{1}{2} f'(c)$ 
A: Let $\varepsilon>0$. Since $f$ is differentiable at $c$, there is some $\delta>0$ such that
$$
\left|\frac{f(c+h)-f(c)}{h}-f'(c)\right|<\varepsilon \quad \forall\, |h| \le \delta.
$$
Since
$$
\frac{1}{n^2}\le\frac{k}{n^2}\le \frac1n \quad \forall k\in \{1,2,\ldots,n\}
$$ 
there is a positive integer $N_\delta$ such that
$$
\frac{1}{n^2}\le\frac{k}{n^2}\le \frac1n\le \delta \quad \forall n\ge N_\delta, \forall k\in \{1,2,\ldots,n\}.
$$
Thus, for every $n\ge N_\delta$ and every $k\in \{1,2,\ldots,n\}$ we have:
$$
\left|\sum_{k=1}^n\frac{k}{n^2}\left[\frac{f\left(c+\frac{k}{n^2}\right)-f(c)}{\frac{k}{n^2}}-f'(c)\right]\right|\le \sum_{k=1}^n\frac{k}{n^2}\varepsilon=\varepsilon\frac{n(n+1)}{2n^2}=\frac12\varepsilon\left(1+\frac1n\right)\le \varepsilon,
$$
i.e.
$$
\lim_{n\to\infty}\left\{\sum_{k=1}^n\left[f\left(c+\frac{k}{n^2}\right)-f(c)\right]-\frac{k}{n^2}f'(c)\right\}=\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2}\left[\frac{f\left(c+\frac{k}{n^2}\right)-f(c)}{\frac{k}{n^2}}-f'(c)\right]=0.
$$
Now, notice that
\begin{eqnarray}
\sum_{k=1}^nf\left(c+\frac{k}{n^2}\right)-nf(c)&=&\sum_{k=1}^n\left[f\left(c+\frac{k}{n^2}\right)-f(c)\right]\\
&=&\sum_{k=1}^n\frac{k}{n^2}\left[\frac{f\left(c+\frac{k}{n^2}\right)-f(c)}{\frac{k}{n^2}}\right]\\
&=&\sum_{k=1}^n\frac{k}{n^2}\left[\frac{f\left(c+\frac{k}{n^2}\right)-f(c)}{\frac{k}{n^2}}-f'(c)\right]+\sum_{k=1}^n\frac{k}{n^2}f'(c)\\
&=&\sum_{k=1}^n\frac{k}{n^2}\left[\frac{f\left(c+\frac{k}{n^2}\right)-f(c)}{\frac{k}{n^2}}-f'(c)\right]+\frac{n(n+1)}{2n^2}f'(c)\\
&=&\sum_{k=1}^n\frac{k}{n^2}\left[\frac{f\left(c+\frac{k}{n^2}\right)-f(c)}{\frac{k}{n^2}}-f'(c)\right]+\frac12\left(1+\frac1n\right)f'(c).
\end{eqnarray}
Hence
\begin{eqnarray}
\lim_{n\to\infty}\left\{\sum_{k=1}^nf\left(c+\frac{k}{n^2}\right)-nf(c)\right\}&=&
\lim_{n\to\infty}\left\{\sum_{k=1}^n\frac{k}{n^2}\left[\frac{f\left(c+\frac{k}{n^2}\right)-f(c)}{\frac{k}{n^2}}-f'(c)\right]+\frac12\left(1+\frac1n\right)f'(c)\right\}\\
&=&\frac12f'(c).
\end{eqnarray}
