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Let $x$ be any positive real number, and define a sequence $\{a_n\}_{n=1}^{\infty}$ by $$ a_n=\frac{[x]+[2x]+\cdots+[nx]}{n^2} $$ where $[x]$ is the largest integer less than or equal to $x$. Prove that $\{a_n\}_{n=1}^{\infty}$ converges to $\frac{x}{2}$.

I'm pretty stuck on this one. Any help would be greatly appreciated.

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  • $\begingroup$ $[kx] = kx - \{kx\}$, where $\{y\}$ denotes the fractional part of $y$. $\endgroup$ – Daniel Fischer Jun 10 '14 at 19:07
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    $\begingroup$ Hint: $t-1\lt[t]\leqslant t$ for every $t$. (The hypothesis that $x$ is positive is not needed.) $\endgroup$ – Did Jun 10 '14 at 19:09
  • $\begingroup$ @Did haha! Thanks! I don't know why I didn't think of that! $\endgroup$ – Mr.Young Jun 10 '14 at 19:15
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I'm sure you've seen the comments by now but notice $$ \begin{array}{ccccc} \frac{x \frac{n(n+1)}{2} - n}{n^2} & < & \frac{[x] + [2x] + \cdots + [nx]}{n^2} & \le & \frac{x \frac{n(n+1)}{2}}{n^2} \\ \downarrow && \Downarrow \text{Squeezed} \Downarrow && \downarrow \\ \frac{x}{2} && \frac{x}{2} && \frac{x}{2} \end{array} $$

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  • $\begingroup$ Both $\frac12n(n-1)$ should read $\frac12n(n+1)$. $\endgroup$ – Did Jun 11 '14 at 7:00
  • $\begingroup$ @Did oops thanks! $\endgroup$ – DanZimm Jun 11 '14 at 10:05
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By definition

$$a_n=\frac{x+2x+\ldots+nx-\{x\}-\ldots\{nx\}}{n^2}=\frac{n(n+1)}{2n^2}x-\frac{\{x\}+\ldots\{nx\}}{n^2}$$

and now just observe that

$$\{x\}+\ldots+\{nx\}\le n$$ and the left hand side is non-negative.

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  • $\begingroup$ I don't think this is sufficient - I believe you really need to squeeze this from both sides. As of right now you have a lower bound, but not something above. $\endgroup$ – DanZimm Jun 10 '14 at 19:19
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    $\begingroup$ Why not sufficient, @DanZimm? The first summand in the right side goes to $\;\frac x2\;$ and the second one is non-negative and less or equal $\;\frac 1n\;$ and thus goes to zero. And you did downvote my answer because "you think" it is not sufficient? $\endgroup$ – Timbuc Jun 10 '14 at 19:24
  • $\begingroup$ Because this isn't a full proof, if you're going to be comparing sequences you need to use the squeeze/sandwich theorem which requires both an upper and lower bound. $\endgroup$ – DanZimm Jun 10 '14 at 19:28
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    $\begingroup$ @DanZimm, this is a full proof. It follows since if $\lim_{n\to\infty} b_n = b$ and $\lim_{n\to\infty} c_n = c$ both exist and $a_n = b_n + c_n$ then $\lim_{n\to\infty} a_n$ exists and is equal to $b + c$. In this case, the first limit converges to $x/2$ and the second to $0$. $\endgroup$ – Joel Jun 10 '14 at 19:30
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    $\begingroup$ @DanZimm, I think Tim's proof is completely correct and complete. What isn't clear to you from it? Besides this, to rush to downvote is not a good policy...ever. And many (most?) answers in this site aren't "full" as most members expect the posters to do some part of the job... $\endgroup$ – DonAntonio Jun 10 '14 at 19:31

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