Calculate the limit: $\lim_{n\rightarrow\infty}\left(\frac{1^{2}+2^{2}+...+n^{2}}{n^{3}}\right)$ 
Calculate the limit: $$\lim_{n\rightarrow\infty}\left(\frac{1^{2}+2^{2}+...+n^{2}}{n^{3}}\right)$$

I'm planning to change the numerator to something else.
I know that $1+2+3+...n = \frac{n(n+1)}{2}$
And now similar just with $2$ as exponent but I did many tries on paper and always failed..
The closest I had is this but it still seems wrong:
$1^{2}+2^{2}+...+n^{2} = \frac{n(n^{2}+1)}{2}$
Well the idea is replacing numerator and then forming it, then easily calculate limit.. But I cannot find the correct thing for numerator..
Any ideas?
 A: $$1+2^2+3^2+...n^2=\frac{n}{6}(n+1)(2n+1)$$
so the limit is
$$\lim_{n\rightarrow\infty}\left(\frac{1}{6}(1+\frac{1}{n})(2+\frac{1}{n})\right)=\frac{2}{6}  = \frac{1}{3}$$
A: Using the identity $\sum_{i=1..n}i^2 = \frac{n(n+1)(2n+1)}{6}$ we can conclude that:
$$\begin{array}{rcl}
\lim_{n\rightarrow\infty}\left(\frac{1^{2}+2^{2}+...+n^{2}}{n^{3}}\right)&=&
\lim_{n\rightarrow\infty}\left(\frac{n(n+1)(2n+1)}{6n^{3}}\right) \\
&=&\lim_{n\rightarrow\infty}\left(\frac{2n^3+\mathcal{O}(n^2)}{6n^3}\right) \\
&=& \frac{1}{3}
\end{array}$$
For the sum of the squares, see this answer: How to get to the formula for the sum of squares of first n numbers?
A: With Stolz-Cesàro:
$$\lim_{n\to\infty}\left(\frac{1^{2}+2^{2}+...+n^{2}}{n^{3}}\right) = \lim_{n\to\infty}\left(\frac{(n+1)^2}{(n+1)^3 - n^3}\right) = \cdots$$
A: From $$\sum_{k=0}^{n}x^{k}=\frac{x^{n+1}-1}{x-1}
 $$ we have, taking the derivative and multiplicand both side by $x$, $$\sum_{k=0}^{n}kx^{k}=\frac{nx^{n+2}-\left(n+1\right)x^{n+1}+x}{\left(x-1\right)^{2}}
 $$ and now taking again the derivative we get $$\sum_{k=0}^{n}k^{2}x^{k-1}=\frac{x^{n}\left(n^{2}\left(x-1\right)^{2}-2n\left(x-1\right)+x+1\right)-x-1}{\left(x-1\right)^{3}}
 $$ and now taking the limit as $x\rightarrow1
 $ we have $$\sum_{k=0}^{n}k^{2}=\sum_{k=1}^{n}k^{2}=\lim_{x\rightarrow1}\frac{x^{n}\left(n^{2}\left(x-1\right)^{2}-2n\left(x-1\right)+x+1\right)-x-1}{\left(x-1\right)^{3}}=\color{green}{\frac{n\left(n+1\right)\left(2n+1\right)}{6}}
 $$ so $$\lim_{n\rightarrow\infty}\frac{\sum_{k=1}^{n}k^{2}}{n^{3}}=\lim_{n\rightarrow\infty}\frac{n\left(n+1\right)\left(2n+1\right)}{6n^{3}}=\color{red}{\frac{1}{3}}.$$
A: You just need to find the leading term in the evaluation of $S_n=1+4+9+\cdots n^2$.
As $S_n-S_{n-1}=n^2$, $S_n$ must be a cubic polynomial, $S_n=an^3+\cdots$ (lower order terms).
Then,
$$an^3-a(n-1)^3=3an^2+\cdots\to n^2,$$ so that $a=\color{green}{1/3}$ and this is the requested limit.

By the same reasoning,
$$\lim_{n\to\infty}\frac{1+2^d+3^d+\cdots n^d}{n^{d+1}}=\frac1{d+1}.$$
This is not surprising, by similarity with the continuous version,
$$\int_0^x x^ddx=\frac{x^{d+1}}{d+1}.$$
It is more interesting to note that the next coefficient is always $1/2$:
$$1+2^d+3^d+\cdots n^d\approx\frac{n^{d+1}}{d+1}+\frac{n^d}2.$$
Indeed,
$$\left(\frac{n^{d+1}}{d+1}+\frac{n^d}2\right)-\left(\frac{(n-1)^{d+1}}{d+1}+\frac{(n-1)^d}2\right)=\\
\frac{n^{d+1}}{d+1}-\frac{n^{d+1}}{d+1}+n^d-\frac{d}2n^{d-1}+\cdots\\
\frac{n^d}2-\frac{n^d}2+d\frac{n^{d-1}}2-\cdots$$ and the terms of degree $d-1$ cancel out.
A: Just for fun, another way. Using Abel's summation we have that $$S_{n}=\sum_{k=1}^{n}k^{2}=\sum_{k=1}^{n}1\cdot k^{2}=n^{3}-2\int_{1}^{n}\left\lfloor t\right\rfloor tdt
 $$ where $\left\lfloor t\right\rfloor 
 $ is the floor function. Since $$t-1\leq\left\lfloor t\right\rfloor \leq t
 $$ we have $$\frac{n^{3}}{3}+\frac{2}{3}\leq S_{n}\leq\frac{n^{3}}{3}+\frac{2}{3}+n^{2}-1
 $$ hence $$\lim_{n\rightarrow\infty}\frac{\sum_{k=1}^{n}k^{2}}{n^{3}}=\frac{1}{3}.$$
A: For variety,
$$\begin{align}
 \lim_{n \to \infty} \frac{1^2 + 2^2 + \ldots + n^2}{n^3}
&=
\lim_{n \to \infty} \frac{1}{n} \left( \left(\frac{1}{n}\right)^2
+ \left(\frac{2}{n}\right)^2 + \ldots + \left(\frac{n}{n}\right)^2
\right)
\\&= \int_0^1 x^2 \mathrm{d}x = \frac{1}{3}
\end{align} 
$$
A: Since you only need the asymptotic behavior, we can rewrite the summand into something easy to manipulate:
$$ x^2 = \frac{x^3 - (x-1)^3}{3} + x - \frac{1}{3}$$
and we can easily compute
$$ \sum_{i=1}^n \frac{i^3 - (i-1)^3}{3} = \frac{n^3}{3} $$
$$ \left| \sum_{i=1}^n i - \frac{1}{3}  \right| < \sum_{i=1}^n n < n^2 $$
This method can be pushed to compute the sum exactly (e.g. as seen the other answers) but an important point is that we don't need to be so exact — to compute the limit we can stop here, since the leftover terms have insignificant sum.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With Stolz-Ces$\grave{\mrm{a}}$ro Theorem:

\begin{align}
\color{#f00}{\lim_{n \to \infty}\pars{1^{2} +2^{2} + \cdots + n^{2} \over n^{3}}} & =
\lim_{n \to \infty}{\pars{n + 1}^{2} \over \pars{n + 1}^{3} - n^{3}} =
\lim_{n \to \infty}{\pars{n + 1}^{2} \over 3n^{2} + 3n + 1}
\\[5mm] & =
\color{#f00}{1 \over 3}\,\lim_{n \to \infty}{\pars{1 + 1/n}^{2} \over
1 + 1/n + 1/\pars{3n^{2}}} = \color{#f00}{1 \over 3}
\end{align}
