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$$\limsup_{n\to\infty}{\lceil{3n^3/5}\rceil-3n^3/5}$$

So, $\limsup$ is basically the supremum of the set consisted of the accumulation points of the expression. I know that $0<\lceil x\rceil-x<1$ but I'm not quite sure what conclusion to draw from this. I've tried watching the values of the expression for $n=1,2,3,4,5$ and it seems the results will be from the set $\{0,1/5,2/5\}$ but since I can't use the calculator on the test it wouldn't be realistic to calculate any values higher than that. It seems kind of reasonable that $2/5$ would be the $\limsup$. Now, I'm not sure how I would go about proving any of this, and also I can't use derivations/integrals and therefore no L'hospital.

Any hint would be greatly appreciated!

Thanks in advance!

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    Hint : $$n^3=5q+0\\n^3=5q+1\\n^3=5q+2\\n^3=5q+3\\n^3=5q+4$$ – 2017-01-27

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Clearly, ${\lceil{3n^3/5}\rceil-3n^3/5}< 1$, and as $3n^3$ is an integer, ${\lceil{3n^3/5}\rceil-3n^3/5} \le 4/5$. Moreover, for $n=5k +3$, $${\lceil{3n^3/5}\rceil-3n^3/5}=4/5$$ So, $$\limsup_{n\to\infty} {\lceil{3n^3/5}\rceil-3n^3/5} = 4/5$$

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    If $n\equiv 3 \pmod 5$ then $3n^3\equiv (3)(27)\equiv 1 \pmod 5.$ You need to take $n\equiv 2 \pmod 5.$ (Possible typo?) – 2017-01-27
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    it's ceiling funtion, not floor. I did the same mistake first, then edited it. – 2017-01-27
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    Sorry .u r right – 2017-01-27
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    Thanks for your answer! I just didn't understand the part that from $3n^3$ being an integer we can deduce that the expression is smaller or equal to 4/5? – 2017-01-27
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    Since $3n^3$ is an integer, $3n^3/5$ will be of the form $k/5$, and since $\lceil 3n^3/5\rceil$ is an integer, the difference will also be of the from $k/5$. Lastly, $k/5<1$ implies $k\le 4$. – 2017-01-27
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    I see, thanks a lot! :) – 2017-01-27