# $x_n=f\left(\frac{1}{n^2}\right)+f\left(\frac{2}{n^2}\right)+\cdots+f\left(\frac{n}{n^2}\right)$, $\lim_{n\to\infty}x_n=\dfrac{f'(0)}{2}$

Let $$f(x)$$ be a function and $$f(0)=0$$, $$f'(0)$$ exists. Let $$x_n=f\left(\frac{1}{n^2}\right)+f\left(\frac{2}{n^2}\right)+\cdots+f\left(\frac{n}{n^2}\right).$$ Prove that $$\lim_{n\to\infty}x_n=\dfrac{f'(0)}{2}$$.

The solution is given by: Since $$f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}$$, we have that $$f(x)=f(0)+f'(0)x+o(x)=f'(0)x+o(x).$$ When $$n\to+\infty$$, we have that $$f\left(\frac{1}{n^2}\right)=\frac{f'(0)}{n^2}+o\left(\frac{1}{n^2}\right)= \frac{f'(0)}{n^2}+o\left(\frac{1}{n}\right),\quad f\left(\frac{2}{n^2}\right)=\frac{2f'(0)}{n^2}+o\left(\frac{2}{n^2}\right)=\frac{2f'(0)}{n^2}+o\left(\frac{1}{n}\right)$$ $$f\left(\frac{k}{n^2}\right)=\frac{kf'(0)}{n^2}+o\left(\frac{k}{n^2}\right)= \frac{kf'(0)}{n^2}+o\left(\frac{1}{n}\right),\dots$$ Hence $$x_n=f'(0)\cdot\frac{1+2+\cdots+n}{n^2}+n\cdot o\left(\frac{1}{n}\right)=f'(0)\cdot\frac{n+1}{2n}+o(1),$$ then $$\lim_{n\to\infty}x_n=\frac{f'(0)}{2}.$$

I have a question: in the above steps, $$o\left(\frac{1}{n}\right)$$ are different, can we do this $$n\cdot o\left(\frac{1}{n}\right)=o(1)$$.

A more accurate question is: suppose there is a two-dimensional sequence $$\{x_{nm}\}_{n,m\in\mathbb{N}}$$ and for a fixed $$m$$, there will always be $$\lim_{n\to\infty}x_{nm}=0$$, is it right that $$\lim_{n\to\infty}\frac{x_{n1}+x_{n2}+\cdots+x_{nn}}{n}=0 ?$$ The original intention of this question is that we used the conclusion $$n\cdot o(1/n)=o(1)$$ in the answers to the above questions. Since these $$o(1/n)$$ are different, set to $$\alpha_{n1},\alpha_{n2},\dots,\alpha_{nn},\quad \alpha_{nk}=o(1/n),\quad 1\leq k\leq n.$$ which is $$\lim_{n\to\infty}\dfrac{\alpha_{nk}}{\dfrac{1}{n}}=\lim_{n\to\infty}n\alpha_{nk}=0,\quad 1\leq k\leq n.$$ Is the limit of the sum of $$\alpha_{n1},\alpha_{n2},\dots,\alpha_{nn}$$ is also $$0$$? That is whether there is $$\lim_{n\to\infty}(\alpha_{n1}+\alpha_{n2}+\cdots+\alpha_{nn})= \lim_{n\to\infty}\frac{n\alpha_{n1}+n\alpha_{n2}+\cdots+n\alpha_{nn}}{n}\stackrel{?}=0.$$ This is the solution:

This is the $$\epsilon$$-$$\delta$$ solution given by a teacher, here we assume $$f'(0)=1$$

• Assuming I haven't made a mistake, if $x_{mn} = \frac{m}{\sqrt{n}}$ then $\lim_{n\to\infty} x_{mn} = 0$ but $\sum_{m} x_{mn} \approx \frac12 n^{3/2}$. As for the original question, I feel like something different is going on or maybe the solution needs to be changed to give more uniform bounds. I need to think some more. Commented Nov 16, 2021 at 0:51
• @TrevorGunn If $\alpha_{nk}=\frac{k}{n\sqrt{n}}$, $\alpha_{nn}$ is not $o(1/n)$. And if $m=n$, then $x_{nn}=\sqrt{n}\not\to 0$ as $n\to \infty$, so your answer maybe not ...
– HGF
Commented Nov 16, 2021 at 1:04
• So, the solution is incorrect/needs more details. Specifically, the solution makes it seem like you just need $o(1/n)$ for constant $m$ whereas we have $o(1/n)$ for $m = n$ as well in the problem. Commented Nov 16, 2021 at 1:10
• @TrevorGunn One answer is to use $\epsilon$- $\delta$ language, see the post answer given by my teacher
– HGF
Commented Nov 16, 2021 at 1:23
• $$\lim_{n\to\infty}n\cdot o\left(\frac{1}{n}\right)= \lim_{n\to\infty}\frac{o\left(\frac{1}{n}\right)}{\left(\frac{1}{n}\right)}=\lim_{t\to 0}\frac{o\left(t\right)}{t}=0.$$ The last step just follows the definition of $o\left(x\right)$ Commented Nov 16, 2021 at 2:35

Here's the $$\varepsilon$$ version basically given by your teacher although I have streamlined it a bit (at least in my mind). I provide this for future purposes in case the original image is deleted or someone searches the specific Tex.

Let $$\varepsilon>0$$ be given. By Taylor's Theorem, there exists a function $$h(x)$$ such that

$$f(x)=f(0)+f'(0)x+h(x)x=f'(0)x+h(x)x$$

and

$$\lim_{x\to 0}h(x)=0$$

Let $$N_1\in\mathbb{N}$$ be given such that $$|h(x)|\leq \frac{\varepsilon}{2}$$ whenever $$|x|\leq \frac{1}{N_1}$$ and let $$N_2\in\mathbb{N}$$ be given by

$$N_2=\left\lceil\frac{|f'(0)|}{\varepsilon}\right\rceil$$

Define $$N=\max\{N_1,N_2\}$$. Then for $$n\geq N$$ we have

\begin{align}\left|x_n-\frac{f'(0)}{2}\right|&=\left|\sum_{i=1}^nf\left(\frac{i}{n^2}\right)-\frac{f'(0)}{2}\right|\\&=\left|f'(0)\sum_{i=1}^n\frac{i}{n^2}+\sum_{i=1}^nh\left(\frac{i}{n^2}\right)\frac{i}{n^2}-\frac{f'(0)}{2}\right|\\&=\left|\frac{f'(0)}{2}+\frac{f'(0)}{2n}+\sum_{i=1}^nh\left(\frac{i}{n^2}\right)\frac{i}{n^2}-\frac{f'(0)}{2}\right|\\&\leq\left|\frac{f'(0)}{2n}\right|+\left|\sum_{i=1}^nh\left(\frac{i}{n^2}\right)\frac{i}{n^2}\right|\end{align}

Then using our definition of $$N$$ we have

$$\leq\frac{\varepsilon}{2}+\left|\sum_{i=1}^n\frac{\varepsilon}{2}\frac{i}{n^2}\right|=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\sum_{i=1}^n\frac{i}{n^2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\sum_{i=1}^n\frac{n}{n^2}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$

and the claim is proved.