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Let $f(x)$ be continuously differentiable on $[0,1]$ and

$$x_n = f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+f\left(\frac{3}{n}\right)+\ldots+f\left(\frac{n-1}{n}\right)$$

Find $$\lim_{n\rightarrow\infty}\left(x_{n+1}-x_n\right)$$



Confusion: I just found a subtle problem in my solution.By the definition of definite integration, like $\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x_i^*)\Delta x$, the sample point $x_i^*$ should be included in its $ith$ subinterval $[x_{i-1},x_i]$. In this case, I am not sure whether the subinterval $[\frac{i-1}{n(n+1)},\frac{i}{n(n+1)}]$ includes the sample point $i$ and $\xi_i$ or not.I can not deduce it from the given condition,

$$\xi_i \in [\frac{i}{n+1},\frac{i}{n}] \implies \frac{i-1}{n(n+1)}\lt \frac{i}{n+1}\lt\xi_i\lt?\lt\frac{i}{n}$$

If those sample points is not always included in the corresponding subinterval, I might not apply the definite integration here. Hope somebody can take a look at this solution.


Update: I rewrite part of my solution and fixed the problem I have before. Thanks very much for all the help!

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3 Answers 3

up vote 4 down vote accepted

Because $f(x)$ is continuously differentiable on $[0,1]$, tben by the mean value theorem:

$$f\left(\frac{i}{n+1}\right)-f\left(\frac{i}{n}\right) = f'(\xi_i)(\frac{i}{n+1} - \frac{i}{n}) \text{ where } \xi_i \in \left[\frac{i}{n+1},\frac{i}{n}\right] \tag{1}$$

Then, by the given formula of $x_n$, we have \begin{align*} \ x_{n+1} - x_n &= \left[f\left(\frac{1}{n+1}\right)-f\left(\frac{1}{n}\right)\right]+\dots +\left[f\left(\frac{n-1}{n+1}\right)-f\left(\frac{n-1}{n}\right)\right]+\left[f\left(\frac{n}{n+1}\right)-f\left(1\right)\right] + f\left(1\right) \\&=f(1) - \sum_{i=1}^{n}i\cdot f'(\xi_i)\left(\frac{1}{n(n+1)}\right) \end{align*}

Hence, by the defintion of definite integration, we have:
\begin{align*} \\\lim_{n\rightarrow\infty}\left(x_{n+1}-x_n\right) &= f(1) - \lim_{n \rightarrow \infty}\sum_{i=1}^{n}i\cdot f'(\xi_i)\left(\frac{1}{n(n+1)}\right) \\&=f(1) - \lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{i}{n+1}\cdot f'(\xi_i)\cdot\frac{1}{n} \\&=f(1) - \int_{0}^{1}xf'(x)dx \space\space\space\dots\text{See Note} \\&=f(1) - \left[xf(x)|_{0}^{1}-\int_{0}^{1}f(x)dx\right] \\ &=\int_{0}^{1}f(x)dx \end{align*}


Note that:: \begin{align*} \\ \sum_{i=1}^{n}i\cdot f'(\xi_i)\left(\frac{1}{n(n+1)}\right) = \sum_{i=1}^{n}\frac{i}{n+1}\cdot f'(\xi_i)\cdot\frac{1}{n} \end{align*} by (1), it is obvious that: $$\frac{i-1}{n} \lt \frac{i}{n+1} \lt \xi_i \lt \frac{i}{n}\space\space\space\text{(where } i \le n) \tag{2}$$ which means $\xi_i, \frac{i}{n+1} \in \left[\frac{i-1}{n},\frac{i}{n}\right]$. Also by $(2)$, we know: $$\sum_{i=1}^{n}\frac{i}{n+1}\cdot f'(\xi_i)\cdot\frac{1}{n} \le \sum_{i=1}^{n}\xi_i\cdot f'(\xi_i)\cdot\frac{1}{n}$$ Because $\frac{i}{n+1}$ and $\xi_i$ share the same subinterval, it implies that if we let $x_{i}^* = \xi_i$(sample points) and $\Delta x=\frac{1}{n}$, then: $$\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{i}{n+1}\cdot f'(\xi_i)\cdot\frac{1}{n}=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\xi_i\cdot f'(\xi_i)\cdot\frac{1}{n}=\int_{0}^{1}xf'(x)dx$$

Hence, it is ok to apply the definition of definite integration here.

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Quite clear. UpVote $0$k. –  Felix Marin Oct 31 '13 at 7:45
    
@FelixMarin I just updated some confusion to my answer, hope you could check that :) –  sundaycat Oct 31 '13 at 8:59
    
$0$k. I'll check tomorrow because it's too late ( 4:30 in the morning ). Thanks. –  Felix Marin Oct 31 '13 at 9:02
    
@FelixMarin Thanks for the help, I rewrite my solution, making it more convincing. Maybe you could take a look at it. –  sundaycat Nov 1 '13 at 0:41
    
TO be fair, you should denote $\xi_{i,n}$, since this element depends both of $i$ and $n$. –  Pedro Tamaroff Nov 1 '13 at 1:16

Heuristically, $x_n \sim n \int_0^1 f$, so $x_{n+1}-x_n \sim \int_0^1 f$ and thus $$\lim_{n \rightarrow \infty} x_{n+1}-x_n= \int_0^1 f$$

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It would be better if you explained $x_n\sim\int_{0}^1f$ in detail. That is almost everything in this problem, all the others are just obvious –  sundaycat Oct 31 '13 at 3:28

Note that: $$\lim_{n\rightarrow\infty}x_n=\lim_{n\rightarrow\infty}\sum_{i=1}^{n-1}f \left(\frac in\right)=\lim_{n\rightarrow\infty}n\cdot\sum_{i=1}^{n-1}f\left(\frac in\right)\frac1n=\lim_{n\rightarrow\infty}n\cdot\sum_{i=1}^{n-1}f\left(x_i\right)\Delta x$$ where $x_i=\frac in\in\left(0,1\right)$ and $\Delta x=\frac1n$.

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But how do you deal with $n$ here? –  sundaycat Oct 31 '13 at 19:01
    
I just show $x_n$ part. Adding $x_{n+1}$ part, $n$ would be canceled. –  Shuchang Nov 1 '13 at 0:39
    
but you can not transform the limit to integration here because you can not get rid of $n$. –  sundaycat Nov 1 '13 at 0:48
    
@sundaycat $\lim_{n\rightarrow\infty}(x_{n+1}-x_n)=\lim_{n\rightarrow\infty}(n+1-n)\sum_{i=‌​1}^{n-1}f(x_i)\Delta x$. Actually $n$ doesn't appear. –  Shuchang Nov 1 '13 at 1:14
    
It seems like a nice way to solve this problem. It would be perfect if you could write down the whole solution here.:) –  sundaycat Nov 1 '13 at 3:34

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