Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 .

share|cite|improve this question
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
up vote 2 down vote accepted

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

share|cite|improve this answer

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

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

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

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.