How to show that $\lim_{n\to\infty}\left\{f(c+\frac{1}{n^2})+f(c+\frac{2}{n^2})+\cdots+f(c+\frac{n}{n^2})-nf(c)\right\}=\frac{1}{2} f'(c)$

Solution 1:

Hint: $f(c+h) = f(c) + f'(c)h +hr(h),$ where $\lim_{h\to 0} r(h) = 0.$

Solution 2:

$\lim_{n\to\infty}\left\{f(c+\frac{1}{n^2})+f(c+\frac{2}{n^2})+\cdots+f(c+\frac{n}{n^2})-nf(c)\right\}=\lim_{n\to\infty}\left\{f(c)+\frac{f'(c)}{n^2}+f(c)+\frac{2f'(c)}{n^2}+...+f(c)+\frac{nf'(c)}{n^2}-nf(c)\right\}= \lim_{n\to\infty}\left\{nf(c)+f'(c)\times\frac{1+2+...+n}{n^2}-nf(c)\right\}= \lim_{n\to\infty}\left\{f'(c)\times\frac{n(n+1)}{2n^2}\right\}= \frac{1}{2} f'(c)$

Solution 3:

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}