Prove that $\{a_n\}_{n=1}^{\infty}$ converges to $\frac{x}{2}$.

Question

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.

$[kx] = kx - {kx}$, where ${y}$ denotes the fractional part of $y$. — Daniel Fischer, Jun 10 '14 at 19:07
Hint: $t-1\lt[t]\leqslant t$ for every $t$. (The hypothesis that $x$ is positive is not needed.) — Did, Jun 10 '14 at 19:09
Does this answer your question? How to show $\lim_{n \to \infty} a_n = \frac{ [x] + [2x] + [3x] + \dotsb + [nx] }{n^2} = x/2$? — Arnaud D., Sep 01 '20 at 14:23

Timbuc · Answer 1 · 2014-06-10T19:34:14.087

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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.

edited Jun 10 '14 at 19:34

answered Jun 10 '14 at 19:15

Timbuc

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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. – DanZimm Jun 10 '14 at 19:19
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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? – Timbuc Jun 10 '14 at 19:24
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. – DanZimm Jun 10 '14 at 19:28
@Dan, are you saying that from $;k_n\ge 0;$ for all $;n;$ and $;k_n\le\frac1n;$ , it doesn't follow that $;k_n\longrightarrow 0;$ ?? Because this is exactly what I showed above... – Timbuc Jun 10 '14 at 19:30
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@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 $0$. – Joel Jun 10 '14 at 19:30
@Joel ah right, nonetheless he didn't justify it in this way (hence my confusion). Silly me! – DanZimm Jun 10 '14 at 19:31
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@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... – DonAntonio Jun 10 '14 at 19:31
@DonAntonio this is true, I have learned from this post. Thankfully I don't often downvote. – DanZimm Jun 10 '14 at 19:32
@Timbuc if you edit your post with something along "and it follows from the linearity of limits" (or just edit it at all) I will change my vote (as of right now I cannot!). – DanZimm Jun 10 '14 at 19:33
Ok @DanZimm, thank you. – Timbuc Jun 10 '14 at 19:33
@Timbuc i just wanted to apologize for the ruckus I had caused before - I was entirely wrong and out of line. – DanZimm Jun 11 '14 at 10:12
@DanZimm, don't worry, please. It happens to us all now and then. – Timbuc Jun 11 '14 at 13:18

DanZimm · Accepted Answer · 2014-06-11T10:05:37.213

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I'm sure you've seen the comments by now but notice $$ \begin{array}{ccccc} \frac{x \frac{n(n+1)}{2} - n}{n^2} & < & \frac{[x] + [2x] + \cdots + [nx]}{n^2} & \le & \frac{x \frac{n(n+1)}{2}}{n^2} \\ \downarrow && \Downarrow \text{Squeezed} \Downarrow && \downarrow \\ \frac{x}{2} && \frac{x}{2} && \frac{x}{2} \end{array} $$

edited Jun 11 '14 at 10:05

answered Jun 10 '14 at 19:14

DanZimm

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Both $\frac12n(n-1)$ should read $\frac12n(n+1)$. – Did Jun 11 '14 at 07:00
@Did oops thanks! – DanZimm Jun 11 '14 at 10:05

Prove that $\{a_n\}_{n=1}^{\infty}$ converges to $\frac{x}{2}$.

2 Answers2