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My work:

$$\lim_{n\to \infty}\sum_{k=1}^n \frac{k^3}{n^4}=\lim_{n\to \infty}\frac1{n^4}\sum_{k=1}^n k^3$$ $$\sum_{k=1}^n k^3 = (1^3+2^3+3^3+...(n-1)^3+n^3)$$ $$\lim_{n\to \infty}\frac{(1^3+2^3+3^3+...(n-1)^3+n^3)}{n^4}$$ $$\lim_{n\to \infty}\frac{1^3+2^3+3^3+...(n-1)^3}{n^4}+\frac1n$$

I know there are some unnecessary terms in my logic, but I thought it might be important to copy down my train of thought in the explanation of my work. I'm not too sure where to progress from here, the only logical step forward would be to prove that $1^3+2^3+3^3+...(n-1)^3$ will never be greater than $n^4$ and therefore converges to zero. Is this logical? Or is there some element to this problem that I am missing?

Kevin McDonough
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1 Answers1

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This is a Riemann sum for $f(t)=t^3$ on the interval $[0,1]$: $$ \sum_{k=1}^n\frac{k^3}{n^4} =\sum_{k=1}^n\left(\frac{k}{n}\right)^3\,\frac1n\xrightarrow[n\to\infty]{}\int_0^1t^3\,dt=\frac14. $$

Another way, without integrals: we know that $$ \sum_{k=1}^n k^3=\frac{n^2(n+1)^2}4. $$ then $$ \sum_{k=1}^n\frac{k^3}{n^4}=\frac{n^2(n+1)^2}{4n^4} =\frac{(1+1/n)^2}4\xrightarrow[n\to\infty]{}\frac14. $$

Martin Argerami
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