Evaluate $\lim_{n \to \infty} \big \{ f(a+1/n^2)+f(a+2/n^2)+\cdots+f(a+n/n^2)-nf(a) \big\}$.

Question

Given $f$ : $\mathbb {R} \to \mathbb {R}$ be differentiable at $x=a$. Evaluate:
$$\lim_{n \to \infty} \big \{ f(a+1/n^2)+f(a+2/n^2)+\cdots+f(a+n/n^2)-nf(a) \big\}.$$
Can I apply the theorem stated: $$\lim_{n\to \infty}\frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n}\right)= \int_0^1 f(x) \, dx$$

Answer 1

Let $\epsilon>0$. Then there exists $\delta>0$ such that $\left|\frac{f(a+h)-f(h)}{h}-f'(a)\right|<\epsilon$ for all $h$ with $|h|<\delta.$ For all $n>\frac1\delta$, we have $0<\frac k{n^2}\le\frac 1n<\delta$ for $1\le k\le n$, hence $$f(a+k/n^2)-f(n)=\frac k{n^2}(f'(a)+ \epsilon_{k,n})\qquad\text{with }|\epsilon_{k,n}|<\epsilon $$ Then $$ \sum_{k=1}^nf(a+k/n^2)\;-nf(a)=\frac{f'(a)\sum_{k=1}^nk}{n^2}+\frac{\sum_kk\epsilon_{k,n}}{n^2}$$ The absolute value of the last term is $<\frac{\sum k}{n^2}\epsilon=\frac{n+1}{2n}\epsilon\le \epsilon$ and the first term is $\frac{n+1}{2n}f'(a)$ and tends to $\frac 12f'(a)$ as $n\to \infty$. We conclude that $$ \left|\left(\sum_{k=1}^nf(a+k/n^2)\;-nf(a)\right)-\frac12f'(a)\right|<2\epsilon\qquad \text{for all }n\gg0.$$ As $\epsilon$ was arbitrary, we cconclude $$ \lim_{n\to\infty}\left(\sum_{k=1}^nf(a+k/n^2)\;-nf(a)\right)=\frac12 f'(a).$$

Remark: Note that we reall yused nothing except differntiability at $a$. Indeed, $f$ may be discontinuous at all other points.