Limit (2)

**NonCommAlg** · June 11th 2009, 03:54 PM

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?

**pankaj** · June 11th 2009, 04:05 PM

Is it Zero

**NonCommAlg** · June 11th 2009, 04:13 PM

Originally Posted by pankaj

Is it Zero

that's the natural candidate and it's absolutely correct. so now we need a proof!

**Random Variable** · June 11th 2009, 04:18 PM

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

**pankaj** · June 11th 2009, 04:56 PM

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

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

What NCA has asked is
$\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^n(-1)^{r-1}f(\frac{r}{n})$
$ =\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})) $

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

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

$ =\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)+...) $

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

$ =0 $

**pankaj** · June 12th 2009, 07:46 AM

I do not believe that I have done it right.

**Bruno J.** · June 20th 2009, 12:45 PM

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!

**NonCommAlg** · June 20th 2009, 03:22 PM

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.

**Bruno J.** · June 20th 2009, 05:45 PM

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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