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Let $\{x_n\}$ be a sequence in $\Bbb R$ and $x_n\to x$ as $n\to \infty$. Then $$\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}\to x$$.

Do anyone know how to solve this kind of problem efficiently? I think I need to estimate $$\left|\frac{(2n-1)x_1+(2n-3)x_2+\dots +3x_{n-1}+x_n}{n^2}-x\right|$$ Is there some other ways to calculate the limit directly?

Thanks for any comments.

VIVID
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Steven Lu
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2 Answers2

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(Fill in the gaps as needed. If you're stuck, explain what you've done and why you're stuck.)

  1. Show that if $ x_i \rightarrow x$, then$ \frac{1}{n} \sum x_i \rightarrow x$.

  2. Show that if $ x_i \rightarrow x$, then $ \frac{1}{n^2} \sum (2i-1) x_i \rightarrow x$.

  3. Hence, conclude that $ \frac{2}{n} \sum x_i - \frac{1}{n^2} \sum (2i-1) x_i \rightarrow x $

Calvin Lin
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  • I'm stuck with the second part at $\frac{1}{n^2}\left| \sum_{k=1}^n (2k-1) (x_k - n^2x)\right|=\frac{1}{n^2} \left| (x_1-nx)+\dots+(2n-1)(x_n-nx) \right|$ – Steven Lu Dec 05 '20 at 06:28
  • @StevenLu Do the same as the first part. Show your work to explain where you're stuck. – Calvin Lin Dec 05 '20 at 06:49
  • $\frac{1}{n^2}\left| \sum_{k=1}^n(2k-1)x_k-n^2x \right|=\frac{1}{n^2}\left| \sum_{k=1}^n(2k-1)(x_k-nx) \right|$, then what should I do to get something like $\frac{1}{n^2}\left|\sum a_k (x_k-x) \right|$? – Steven Lu Dec 05 '20 at 11:23
  • @StevenLu In the linked solution, they broke it up into 2 parts. The first part had a finite bound, and the second part was bounded by $(x_k-x)$. Do the same here. (Note, there are several mistakes in your work. E.g. do you want $(x_k - nx)$? Also please define terms when you use them. E.g. what is $a_k$?) – Calvin Lin Dec 05 '20 at 11:40
  • Thanks for your hint and not write down the answer, I finally realize how to do it. For this kind of problem, what should I do first then I can solve it easy? Should I break the summation into. several parts, just as you did, and then try to find the limit? – Steven Lu Dec 05 '20 at 12:04
  • @StevenLu The idea is that you want to find ways to bound the terms. So you break it up into manageable parts, and use as many as you need to. Dealing with the "initial sequence" and "final sequence" made a lot of sense here. In some cases, you might also want to separate out the "middle sequence", esp by considering the coefficients. E.g. try showing that $ \frac{1}{2^n} \sum { n \choose i } x_i \rightarrow x$ (I've not spent much time thinking about this, and so this might not be true, but I'm guessing that it is) – Calvin Lin Dec 05 '20 at 12:11
  • I'm not quite sure how to exactly separate out the middle sequence, but I have think about it. Since $\sum_{k=0}^n{n \choose k}=2^n$, we can write $\frac{1}{2^n}\left|\sum_{k=0}^n{n \choose k}(x_k-x) \right|$. And I think exponential go faster than polynomial, all coefficient terms $\frac{n \choose k}{2^n}$ will tend to $0$, and since $x_n \to x$, I think we may separate this sequence into tow parts as previous one. – Steven Lu Dec 05 '20 at 13:31
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This limit problem looks very much like intentionally designed for the use of the Stolz-Cesàro theorem:

So, let $$a_n =\sum_{k=1}^n(2n-(2k-1))x_k = \sum_{k=1}^n(2(n-k)+1)x_k$$ and $$b_n = n^2$$

So, $$a_{n+1}-a_n = \sum_{k=1}^{n+1}(2(n+1-k)+1)x_k - \sum_{k=1}^n(2(n-k)+1)x_k$$$$ = x_{n+1}+2\sum_{k=1}^n x_k$$

Hence, $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}= \frac{x_{n+1}+2\sum_{k=1}^n x_k}{2n}=\underbrace{\frac{x_{n+1}}{2n}}_{\stackrel{n\to\infty}{\longrightarrow}0} + \underbrace{\frac{\sum_{k=1}^n x_k}{n}}_{\stackrel{n\to\infty}{\longrightarrow}x}$$

where the last limit is just another application of Stolz-Cesàro.

trancelocation
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