2
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • $\begingroup$ $[kx] = kx - \{kx\}$, where $\{y\}$ denotes the fractional part of $y$. $\endgroup$ – Daniel Fischer♦ Jun 10 '14 at 19:07
  • 7
    $\begingroup$ 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
4
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 3
    $\begingroup$ @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
  • 3
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.