Find the limit: $\lim_{n\to\infty} \frac{ \sum_{i=1}^n\lfloor i^3x \rfloor}{n^4}$ $$\lim_{n \to {\infty}} \frac{ \sum_{i=1}^n\lfloor i^3x \rfloor}{ n^4}$$
My work
$$\lim_{n \to {\infty}} \frac{ \sum_{i=1}^ni^3x}{ n^4} -\lim_{n \to {\infty}} \frac{ \sum_{i=1}^n{\{i^3x\}}}{ n^4}$$
$$\lim_{n \to {\infty}} \frac{x{((n)(n+1))}^2}{4 {n^4}}-\lim_{n \to {\infty}}\frac{ \sum_{i=1}^n{\{i^3x\}}}{ n^4}$$
{.} represent fractional part
Can I say this
 $$\lim_{n \to {\infty}}\frac{ \sum_{i=1}^n{\{i^3x\}}}{ n^4} = 0$$ 
 A: Yes, it is true that
$$\lim_{n \to {\infty}}\frac{ \sum_{i=1}^n{\{i^3x\}}}{ n^4} = 0$$
since $0\le\{y\}<1$ for any $y\in\mathbb R$.
Alternatively, we can use a chain of inequalities for the floor function
$$\sum_{i=1}^n(i^3x-1)<\sum_{i=1}^n\lfloor i^3x \rfloor\le\sum_{i=1}^ni^3x$$
and approximate the sum by definite integrals
$$
\frac{(n-1)^4}4-\frac14=\biggl[\frac{x^4}{4}\biggr]_1^{n-1}=\int_1^{n-1}x^3\mathrm dx\le\sum_{i=1}^ni^3\le\int_0^nx^3\mathrm dx=\biggl[\frac{x^4}{4}\biggr]_0^n=\frac{n^4}4
$$
to obtain the answer.
A: The approach you have chosen is good and probably the most straightforward for this limit. Just in case it will be useful for you in the future, I will mention that you can also use Stolz-Cesaro theorem to get
$$\lim\limits_{n\to\infty} \frac{\sum_{i=1}^n \lfloor i^3x \rfloor}{n^4} = 
\lim\limits_{n\to\infty} \frac{\lfloor (n+1)^3x\rfloor}{(n+1)^4-n^4} = 
\lim\limits_{n\to\infty} \frac{\lfloor (n+1)^3x\rfloor}{4n^3+6n^2+4n+1}.$$
Of course, we should also check that the assumptions of Stolz-Cesaro theorem are satisfied.
