I am trying to find the limit: $$\lim_{n\to\infty}\left(\lim_{k\to\infty}\int_{0}^{1}\left ( \frac{[kx]}{k} \right )^{n}\text dx\right)$$

I have suggested a solution below. But I feel there can be other ways to directly observe that the answer is zero. Please suggest (preferably easier) ways to find the values of the integral other than mine.

My answer: We try to calculate the innermost integral by breaking the integral from $0 \space\text{to} \space \frac 1k$,$\frac 1k\space \text{to}\space \frac 2k $ and so on, $$\int_{0}^{1}\left ( \frac{[kx]}{k} \right )^{n}dx=\int_{0}^{\frac 1k}0^n\cdot dx+\int_{\frac 1k}^{\frac 2k}(\frac 1k)^n \cdot dx+\space...\space+\int_{\frac{k-1}{k}}^{1}(\frac{k-1}{k})^ndx$$ $$=\frac 1{k^{n+1}}(0+1^n+2^n+3^n+\space...\space+(k-1)^n)$$ $\implies $$$\lim_{n\to\infty}(\lim_{k\to\infty}\frac{\sum_{i=1}^{k-1}i^n}{k^{n+1}})$$ Now we see that the sum inside the bracket inside the bracket $ <\space \frac{(k-1)^{n}}{k^{n+1}}\cdot k < \frac{(k-1)^n}{k^n} < 1$ . so when $\lim_{n\to \infty}$, the answer $ =0$.

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    $\begingroup$ Uhhh...what way is yours? $\endgroup$
    – Ron Gordon
    Feb 28, 2018 at 22:09
  • $\begingroup$ I have a suggested answer. Sorry to not mention it. $\endgroup$
    – sonu
    Mar 1, 2018 at 4:38
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    $\begingroup$ I actually do not see any problem with putting a complete answer in an answer-post instead of the question-post, as long as the question-post mentions it. The SE system explicitly encourages self-answering as a way of sharing knowledge. So I think next time you can do that and not delete your answer. $\endgroup$
    – user21820
    Mar 14, 2018 at 14:29
  • $\begingroup$ Yes I didn't delete the answer purposefully but Mr. John Ma edited my question having the answer inside so I didn't mind much and deleted the answer. $\endgroup$
    – sonu
    Mar 14, 2018 at 15:50

1 Answer 1


The innermost integral is non-negative but bounded by $$ \int_{0}^{1}\left(\frac{kx}{k}\right)^{n}\,dx = \frac{1}{n+1}.$$


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