What does the expression tend to? Let a function defined on the real line be differentiable at the real number $1$. As the integer tends to infinity, what does the expression $f (1+1/n^2)+f(1+2/n^2)+⋅⋅⋅f(1+n/n^2)- nf(1)$ tend to?
My answer is $0$, since each term gets multiplied by $0$. I got this by distributing the $nf(1)$ between the other $n$ terms. And, since the the function is differentiable at $1$, it is a bounded quantity which multiplied by $0$ each time. Is my answer correct?
 A: Let's assume that the first derivative is continuous at $1$,  and apply the mean value theorem: 
$$f(1+a) = f(1) + a f'(c),$$ 
where $c$ is between $1$ and $1+a$. Then we have 
$$ \mbox{sum} = n f(1) + \sum_{j=1}^n \frac{j}{n^2} f'(c_j)-n f(1) = \frac{1}{n^2} \sum_{j=1}^n j f'(c_j)=(*)$$ 
Now
$f'(c_j) = f'(1) + f'(c_j) - f'(1)$. Therefore, 
$$(*) = \frac{f'(1)}{n^2}\sum_{j=1}^n j + R_n,$$ 
where $$R_n = \frac{1}{n^2} \sum_{j=1}^n j \left(f'(c_j) - f'(1)\right).$$ 
Now, 
$$\sum_{j=1}^n j = \frac{(1+n)n}{2},$$
and as for $R_n$, or each $j$, $|c_j-1| \le 1/n$, so it follows that $$|f'(c_j)-f'(1)|\le \epsilon_n = \sup_{1<x<1+1/n} |f'(x) -f'(1)|.$$ 
Since $f'$ is continuous at $1$, $\epsilon_n\to 0$. Thus, 
$$|R_n|< \epsilon_n \frac{(1+n)n}{2n^2}\to 0,\mbox{ as }n\to\infty.$$
It follows that 
$$\lim_{n\to\infty} (*) = \frac{f'(1)}{2}.$$ 
A: No, your answer is not correct. You basically said that" $\sum_{n=0}^{\infty} a_n = 0$ since $\lim\limits_{n \rightarrow 0} a_n = 0$"; which is false. 
One way to do this is to use the Taylor series (you mention that the function is differentiable, that should probably be used):
$\frac{n^2}{i}(f(1+\frac{i}{n^2})-f(1)) = f'(1) + o(\frac{i}{n^2}) $
$f(1+\frac{i}{n^2})-f(1) = \frac{i}{n^2}f'(1) + o(\frac{i^2}{n^4}) = \frac{i}{n^2}f'(1) + o(\frac{1}{n^2}) $
$\sum_{i=1}^n  f(1+\frac{i}{n^2})-f(1) = (\sum_{i=1}^n \frac{i}{n^2}f'(1)) + o(\frac{1}{n}) $
So it has the same limit as $ (f'(1)\sum_{i=1}^n \frac{i}{n^2}) = f'(1)\frac{n(n+1)}{2n^2}$
I let you finish
