Why are the sum of squares afraid of one cube?

Using $\ 1^{2} +2^{2} +3^{3} +...+ n^{2} = \frac{n(n+1)(2n+1)}{6}$ : $\lim_{n \rightarrow\infty } \frac{1^2 +2^2 +3^2 +...+ n^2 }{n^3} = \lim_{n \rightarrow\infty } \frac{n(n+1)(2n+1)}{6n^3}$ $= \lim_{n \rightarrow\infty } \frac{2n^2 +3n+1}{6n^2}$ $= \frac{2n^2}{6n^2} = \frac{1}{3} \therefore\boxed{a+b = 4}$ Note that I got rid of the smaller powers of ${n}$ because they become insignificant compared to $\ n^{2}$ as ${n}$ tends to infinity.

$1^2$ + $2^2$ + $3^2$ +....+ $n^2$ = $\frac{n}{6}$ (n+1)(2n+1).

$\frac{n}{6}$ (n+1)(2n+1)/ $n^3$ = $\frac{1}{6}$ (1)(1+ $\frac{1}{n}$ (2+ $\frac{1}{n}$ ). When n tends to infinity, both $\frac{1}{n}$ and $\frac{2}{n}$ tend to 0 so we have $\frac{2}{6}$ = $\frac{1}{3}$ as the limit. Now a+b = 1+3 = $\boxed{4}$

$\frac{\sum_{k=1}^{n} k^2}{n^3} = \frac{n(n+1)(2n+1)}{6n^3} \\ = \frac{2n^2 + 3n + 1}{6n^2} \\ \lim_{n\to\infty} \frac{2n^2 + 3n + 1}{6n^2} = \frac{\infty}{\infty}$ Apply L'Hôpital's rule twice: $\lim_{n\to\infty} \frac{2n^2 + 3n + 1}{6n^2} \\ = \lim_{n\to\infty} \frac{4n + 3}{12n} \\ = \lim_{n\to\infty} \frac{4}{12} \\ = \frac{1}{3} \\ 1+3=\boxed{4}$

$\lim_{n\rightarrow\infty}\frac{1^2+2^2+3^2+\ldots+n^2}{n^3}=\lim_{n\rightarrow\infty}\frac{1}{n}\displaystyle\sum_{k=1}^{n}\left(\frac{k}{n}\right)^2$ $=\displaystyle\int_{0}^{1}x^2\mathrm dx=\frac{1}{3}$ $\therefore\boxed{ a+b=4}$

I posted this same problem before. This came from the Piskunov calculus book, I think.