# Finding the limit .

Consider $f$ is differentiable,

$$\lim_{n\to \infty} \frac{1}{n^2} \sum_{k=1}^n \frac{f(a+\frac{k}{n^2}) -f(a)}{\frac{k}{n^2}}$$ . My idea was , Since $f$ is differentiable each term in the sum exists $\forall n$ , hence say $M$ be the max so we have

$$\lim_{n\to \infty} \frac{1}{n^2} n.|M|$$ Hence the limit is $0$. Can you guys help me out .

-
Is it not $\int_a^af'(a)\,dx$?? – Jp McCarthy Dec 2 '13 at 13:35
@JpMcCarthy : It looks like that , but can you give a slight explanation how we can deduce the expression using rieman sum . – Complex analysis Dec 2 '13 at 13:39
No I cannot to be honest with you --- I would have to play a bit fast and loose... looking below I see a proper answer and that it should have been $\displaystyle \frac1n \int_a^af'(a)\,dx$ perhaps. – Jp McCarthy Dec 2 '13 at 13:53

Because $f$ is differentiable, there limit $$\frac{f(a+t)-f(a)}{t}$$ exists and is equal to $f'(a)$. Thus, there is a value $t_0$ such that for $t < t_0$ it holds that $\frac{f(a+t)-f(a)}{t} \in (f'(a)-1,f'(a)+1)$. In any case, for sufficiently small $t$, $|\frac{f(a+t)-f(a)}{t}| < |f'(a)|+1$. On the other hand, if you select $F:= \max_{a\leq x \leq x+1} |f(x)|$ then for $t_0 < t < 1$ you have $|\frac{f(a+t)-f(a)}{t}| < F/t_0$. If $M = \max(|f'(a)|+1,F/t_0)$, then $|\frac{f(a+t)-f(a)}{t}| < M$ for all $t \leq 1$.

Applying this to the sum in question, we get

$$\sum_{k=1}^n \frac{f(a+\frac{k}{n^2}) -f(a)}{\frac{k}{n^2}} < n M.$$ Thus, $$\lim_{n\to \infty} \frac{1}{n^2} \sum_{k=1}^n \frac{f(a+\frac{k}{n^2}) -f(a)}{\frac{k}{n^2}} \leq \lim_{n\to \infty}\frac{M}{n} =0.$$

-

Note that, since f is differentiable, then we can approximate (when $n$ is large ) as

$$f(a+\frac{k}{n^2})\sim f(a)+f'(a)\frac{k}{n^2}.$$

Now, we have

$$\frac{1}{n^2} \sum_{k=1}^n \frac{f(a+\frac{k}{n^2}) -f(a)}{\frac{k}{n^2}} \sim \frac{1}{n^2} \sum_{k=1}^n \frac{(f(a)+f'(a)\frac{k}{n^2})) -f(a)}{\frac{k}{n^2}}$$

$$=\frac{f'(a)}{n^2}\sum_{k=1}^{n}1=\frac{f'(a)}{n^2}n=\frac{f'(a)}{n}\longrightarrow_{n\to \infty} 0.$$

-

Maybe you could use $\frac{f(a+\frac{k}{n^2})-f(a)}{\frac{k}{n^2}} = f'(\xi)$ where $\xi\in (a, a+\frac{k}{n^2})$

So $\frac{1}{n^2}\sum\limits_{k=1}^n\frac{f(a+\frac{k}{n^2})-f(a)}{\frac{k}{n^2}} = \frac{1}{n^2}\sum\limits_{k=1}^nf'(\xi_k) \sim \frac{f'(a)}{n} \sim 0$

-
So that means the limit is $0$ right ? – Complex analysis Dec 2 '13 at 13:40
@Complexanalysis, I got 0 as an answer – tenpercent Dec 2 '13 at 13:45