# Thread:

1. ## Limit (2)

Let $f: [0,1] \longrightarrow \mathbb{R}$ be continuous. We all know that $\lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n f \left(\frac{j}{n} \right)=\int_0^1 f(x) \ dx.$ Now let $L=\lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n (-1)^{j-1}f \left(\frac{j}{n} \right).$ What do you think the value of $L$ is?

2. Is it Zero

3. Originally Posted by pankaj
Is it Zero
that's the natural candidate and it's absolutely correct. so now we need a proof!

4. Can you direct me to a proof of the limit that everyone knows except for me?

5. $\int_{a}^{b}f(x)dx=\lim_{h\to 0}h(f(a+h)+f(a+2h)+f(a+3h)+.....+f(a+nh))$

If a=0 and b=1,we get $h=\frac{b-a}{n}=\frac{1-0}{n}=\frac{1}{n}.$

It is then that we have $\int_{0}^{1}f(x)dx=\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^nf(\frac{r}{n})$= $\lim_{h\to 0}h(f(h)+f(2h)+f(3h)+.....+f(nh))$

$\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^n(-1)^{r-1}f(\frac{r}{n})$
$<br /> =\lim_{n\to \infty}\frac{1}{n}(f(\frac{1}{n})-f(\frac{2}{n})+f(\frac{3}{n})-..........+(-1)^{n-1}f(\frac{n}{n}))<br />$

$<br /> =\lim_{h\to 0}h(f(h)-f(2h)+f(3h)-f(4h)+f(5h)-....+(-1)^{n-1}f(nh))<br />$

= $\lim_{h\to 0}h(f(h)+f(3h)+f(5h)+....)-\lim_{h\to 0}h(f(2h)+f(4h)+f(6h)+...)$

$<br /> =\frac{1}{2}\lim_{h\to 0}(2h)(f(h)+f(3h)+f(5h)+....)-\frac{1}{2}\lim_{h\to 0}(2h)(f(2h)+f(4h)+f(6h)+...)<br />$

$<br /> =\frac{1}{2}\int_{0}^1f(x)dx-\frac{1}{2}\int_{0}^1f(x)dx<br />$

$<br /> =0<br />$

6. I do not believe that I have done it right.

7. Originally Posted by NonCommAlg
Let $f: [0,1] \longrightarrow \mathbb{R}$ be continuous. We all know that $\lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n f \left(\frac{j}{n} \right)=\int_0^1 f(x) \ dx.$ Now let $L=\lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n (-1)^{j-1}f \left(\frac{j}{n} \right).$ What do you think the value of $L$ is?
Write

$L_n = \frac{1}{n} \sum_{j=1}^n (-1)^{j-1}f \left(\frac{j}{n} \right)$. Then

$L_{2n} = \frac{1}{2n} \sum_{j=1}^n \Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n} \right)\Big]$

hence
$L_{2n} \leq \frac{1}{2n} n \mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n}\right)\Big] = \frac{1}{2}\mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n}\right)\Big]$

and similarily $L_{2n} \geq \frac{1}{2}\mbox{ inf }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n}\right)\Big]$ so that

$\frac{1}{2}\mbox{ inf }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n}\right)\Big] \leq L_{2n} \leq \frac{1}{2}\mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-f\left(\frac{2j}{n}\right)\Big]$

but both sides go to 0 as $n \rightarrow \infty$ since $f$ is continous!

8. Originally Posted by Bruno J.
Write

$L_n = \frac{1}{n} \sum_{j=1}^n (-1)^{j-1}f \left(\frac{j}{n} \right)$. Then

$L_{2n} = \frac{1}{n} \sum_{j=1}^n \Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n} \right)\Big]$ n here should be 2n.

hence
$L_{2n} \leq \frac{1}{n} n \mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n}\right)\Big] = \mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n}\right)\Big]$

and similarily $L_{2n} \geq \mbox{ inf }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n}\right)\Big]$ so that

$\mbox{ inf }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n}\right)\Big] \leq L_{2n} \leq \mbox{ sup }_{1 \leq j \leq n}\Big[f \left(\frac{2j-1}{n} \right)-\left(\frac{2j}{n}\right)\Big]$

but both sides go to 0 as $n \rightarrow \infty$ since $f$ is continous!

even $L_{2n} \to 0$ will not necessarily imply that $L_n \to 0$ because we don't know that $\lim L_n$ exists.

9. Thanks for the heads up! I was in a hurry when I wrote that (excuses excuses ).

Indeed $L_{2n}\rightarrow 0$ does not imply the result, but a slight modification of the argument should show that $L_{n}\rightarrow 0$, ex. by writing

$L_{2n+1} = \frac{1}{2n+1}\Big[\sum_{j=1}^n\Big[f\Big(\frac{2j-1}{n}\Big)-f\Big(\frac{2j}{n}\Big)\Big]+f\Big(\frac{2n+1}{n}\Big)\Big]$
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