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If $[$.$]$ denotes the greatest integer function, then find the value of $\lim_{n \to \infty} \frac{[x] + [2x] + [3x] + … + [nx]}{n^2}$

What I did was, I wrote each greatest integer function $[x]$ as $x - \{x\}$, where $\{.\}$ is the fractional part. Hence, you get

$\lim_{n \to \infty} \frac{\frac{n(n+1)}{2}(x-\{x\})}{n^2}$

The limit should then evaluate to $\frac{x-\{x\}}{2}$

But the answer given is $\frac{x}{2}$. What am I missing here?

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$$\frac{\lfloor x\rfloor+\ldots+\lfloor nx\rfloor}{n^2}=\frac{x+2x+\ldots nx-\{x\}-\ldots-\{nx\}}{n^2}=$$

$$=\frac{n(n+1)}{2n^2}x-\frac{\{x\}+\ldots+\{nx\}}{n^2}\xrightarrow[n\to\infty]{}\frac12x-0=\frac x2$$

since the second addend above tends to zero:

$$\frac{\{x\}+\ldots+\{nx\}}{n^2}\le\frac n{n^2}=\frac1n\xrightarrow[n\to\infty]{}0$$

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  • $\begingroup$ I didn't really get how the second addend (fractional part thing) tends to zero, as n tends to infinity. Isn't it in an inderminate form as it is? Also, what has been done in the last line of the answer? $\endgroup$ – skb May 28 '18 at 20:20
  • $\begingroup$ @skb Every term $\;\{kx\}\;$ is less than $\;1\;$ , so that whole sum's numerator is less that $\;1+1+\ldots+1=n\;$ ... $\endgroup$ – DonAntonio May 28 '18 at 20:22
  • $\begingroup$ Have you used the sandwich theorem in the last line? But shouldn't there be another expression less than it for it to work? $\endgroup$ – skb May 28 '18 at 20:24
  • $\begingroup$ @skb But isn't it obvious that the whole expression is greater than zero or equal to it? You can complete that argument... $\endgroup$ – DonAntonio May 28 '18 at 20:27
  • $\begingroup$ Got it. Thanks a lot for the help! $\endgroup$ – skb May 28 '18 at 20:27
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Note that by Stolz-Cesaro

$$\lim_{n \to \infty} \frac{\lfloor x\rfloor + \lfloor 2x \rfloor + \lfloor3x\rfloor + … + \lfloor nx]}{n^2}=\lim_{n \to \infty} \frac{\lfloor(n+1)x\rfloor}{(n+1)^2-n^2}=\lim_{n \to \infty} \frac{\lfloor(n+1)x\rfloor}{2n+1}=$$ $$=\lim_{n \to \infty} \frac{(n+1)x}{2n+1}-\frac{\{(n+1)x\}}{2n+1}$$

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