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Find the limit of $\lim_{n\rightarrow ~0}\frac{(x)+(2x)+\cdots (nx)}{n^2}$, where, $(x)=x-[x]$ and $[x] $ is the greatest integer function(the fractional part function).

I feel, as $n \rightarrow 0$ this limit goes to infinity, but the options given are $x~,~x/2,~x/3,~x/4$.

How this is happening, I double checked the question paper, in question $n$. is tending to 0 only not to $\infty$.

I found a similar question here

Martin Sleziak
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sabeelmsk
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    I think this is a typo in the question as $n$ needs to be a natural number. Either they meant $n \rightarrow \infty$ or $x \rightarrow 0$. – quarague Sep 19 '19 at 11:03
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    The limit has got to be $n\to\infty$. The reason is because, if $n<1$, say $n=\frac1{100}$, how do you make sense of $$ (x)+(2x)+\cdots (nx)? $$ Or for that matter, $$ a_1 + a_2 + \dots + a_n?$$ this notation implies that $n$ is an integer. There is a notion of $n\to 0$ along integers but I doubt its what the question meant – Calvin Khor Sep 19 '19 at 11:05
  • I would say that the limit for n to 0 is nonsense in this case! – georg Sep 19 '19 at 11:06
  • In addition, the answer for integer values of $x$ is 0, as $(k)=0$ for every integer $k$. So there needs to be more information on $x$ On the other hand, the limit as $n\to\infty$ is clearly 0 as the top is then bounded by $n$ and $n/n^2 \to 0$..... – Calvin Khor Sep 19 '19 at 11:10
  • Yes, the question is quite wonky indeed. If we work with the sequence as $\sum_{k = 0}^n kx \over n^2$ we may get an answer but it’s quite an uncomfortable thing to do. – Certainly not a dog Sep 19 '19 at 11:11
  • The answer if $\frac x 2$ for that one. @Certainlynotadog – nicomezi Sep 19 '19 at 11:21
  • Yes, it seems to be @nicomezi. (I had made a silly mistake in my original comment) – Certainly not a dog Sep 19 '19 at 11:40
  • Note: if the choice of answers is $x, x/2, x/3, x/4$ then $(x)$ is not the fractional part. It is just ordinary parenthesis, and the answer is $x/2$. To do this you have to know $(1+2+\dots+n) = n(n-1)/2$. – GEdgar Sep 20 '19 at 13:01
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    If $(x)$ is fractional part, then it is always in $[0,1]$ and the numerator is at most $n$, so with denominator $n^2$ the limit is zero. – GEdgar Sep 20 '19 at 13:03

3 Answers3

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This was too long for a comment, hopefully this helps:

Using $1+2+\dots+n=\frac{n(n+1)}{2}$ we find that: \begin{align*} \lim_{n\rightarrow\infty} \frac{(x)+(2x)+\dots+(nx)}{n^2} &=\lim_{n\rightarrow\infty} \frac{x-[x]+2x-[2x]+\dots+nx-[nx]}{n^2}\\ &=\lim_{n\rightarrow\infty} \frac{x(1+2+\dots+n)}{n^2}-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\ &=x\lim_{n\rightarrow\infty} \frac{n(n+1)}{2n^2}-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\ &=\frac x2-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\ \end{align*}

Cedric Brendel
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    The last line is harder to compute then where you came from. But the result, $\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}=\frac x 2$ (you forgot to wrote the $\frac 1 2$ in the last line), is pretty interesting to me. – nicomezi Sep 19 '19 at 11:46
  • Corrected, thanks :) Do you really think so? – Cedric Brendel Sep 19 '19 at 11:48
  • Using $[kx]=k[x]+[k{x}]$ for $k\in\mathbb{Z}$ we also find that $$\lim_{n\rightarrow\infty} \frac{[{x}]+[2{x}]+\dots+[n{x}]}{n^2}=\frac{{x}}{2}$$ – Cedric Brendel Sep 19 '19 at 11:57
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    Yeah I mean it, since I first thought it was false. So at least I did not expect it. – nicomezi Sep 19 '19 at 12:04
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For $n \longmapsto 0$ the nominator is meaningless!

For $n \longmapsto +\infty$ we have: $$\forall_y:0 \leqslant y-[y] <1 \Longrightarrow 0 \leqslant (y) <1$$ $$\lim_{n \longmapsto +\infty} \frac{(x)+...+(nx)}{n^2} = \lim_{n \longmapsto +\infty} \frac{(x)+...+(nx)}{n}.\frac{1}{n} =$$ $$= \lim_{n \longmapsto +\infty} \frac{Mean((x),...,(nx))}{n} = \lim_{n \longmapsto +\infty} \frac{Bounded}{n} = 0$$ In fact the mean of (ix)'s is stays in $[0,1)$ since all of them are in that area.

Ali Ashja'
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    Would have been easier to just say that $(nx)<1$, for all positive $n,x$. Plus, you are using this fact to deduce that the mean is bounded, aren't you ? – nicomezi Sep 19 '19 at 11:23
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not an answer to the stated question (we use denominator $n$, not $n^2$) but interesting nonetheless

See information on equidistributed sequences ...

If $x$ is irrational, then the sequence $(x), (2x), (3x), \dots$ of fractional parts is equidistributed on $[0,1]$. Therefore, for every Riemann integrable function on $[0,1]$ we have $$ \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n f\big((kn)\big) = \int_0^1 f(t)\;dt . $$ The function $f(t) = t$ is Riemann integrable, so $$ \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n (kn) = \int_0^1 t\;dt =\frac{1}{2}. $$ if $x$ is irrational. If $x$ is rational, this limit could be something other than $1/2$.

GEdgar
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