Evaluate $\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}.$ 
Evaluate: $$\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}.$$

I'm learning limits for the first time and this is an exercise problem from my book. Here is my solution to the problem:
Let $S=1\cdot3+2\cdot4+\dots+n(n+2)\\ =(1^2+2)+(2^2+4)+\dots+(n^2+2n)\\ =(1^2+2^2+\dots+n^2)+2(1+2+\dots+n)\\ =\frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}2\\ =\frac13n^3+\frac32n^2+\frac76n$
Hence, $\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}\\ =\lim\limits_{n\to\infty}\frac13+\frac3{2n}+\frac7{6n^2}\\ =\frac13.$
I'm quite sure about the solution. But my book says the answer is $\frac16$. So, is the answer in the book wrong, or am I missing something? And can the problem be solved with L'Hôpital's rule? (I've just started learning the rule and I don't know how to solve this using this). Some other methods to solve the problem are also welcome.
 A: Yet another solution, using Stolz Cesaro theorem, which states that if $(b_n)_{n \in \mathbb N_+}$ is an increasing sequence diverging to $+\infty$ and if limit $\lim_{n \to \infty} \frac{a_n-a_{n-1}}{b_n - b_{n-1}}$ exists, then $\lim_n \frac{a_n}{b_n} = \lim_n \frac{a_n - a_{n-1}}{b_n-b_{n-1}}$.
Using this with $b_n = n^3$ and $a_n = 1\cdot 3 + ... + n \cdot (n+2)$ you arrive at $$ \frac{a_n - a_{n-1}}{b_n - b_{n-1}} = \frac{n(n+2)}{n^3 - (n-1)^3} = \frac{n(n+2)}{n^3-n^3+3n^2 - 3n + 1} = \frac{n^2+2n}{3n^2-3n+1} \xrightarrow[n \to \infty]{}\frac{1}{3}$$
A: $$ \lim_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3} = \lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{k(k+2)}{n^3}\right) = \lim_{n\to\infty}\left(\frac{1}{n^3}\ \sum_{k=1}^{n}k^2+2k\right)$$
$$ = \lim_{n\to\infty}\left(\frac{1}{n^3}\left(\frac{n(n+1)(2n+1)}{6}+n(n+1)\right)\right) = \frac{1}{3}$$
A: Your answer is right. You could also use the Stolz-Cesàro theorem, a discrete analog to the L'Hopital rule that I think should be better known:
$$\lim_{n\to\infty}\frac{\sum_{k=1}^nk(k+2)}{n^3} = \lim_{n\to\infty}\frac{(n+1)(n+3)}{(n+1)^3-n^3} = \lim_{n\to\infty}\frac{n^2+4n+3}{3n^2+3n+1} = \frac{1}{3}$$
A: Your answer is indeed correct. Here is just another way of solving
$$\begin{align}
\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^n k(k+2) 
&= \lim_{n\to\infty}\left(\color{blue}{\frac{1}{n}\sum_{k=1}^{n}\frac{k^2}{n^2}} + \color{fuchsia}{\frac{1}{n^3}\sum_{k=1}^n 2k}\right) \\[1mm]
&= \color{blue}{\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k^2}{n^2}} + \color{fuchsia}{\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^n 2k} \\[1mm]
&= \color{blue}{\int_0^1 x^2dx} + \color{fuchsia}{0} \\[1mm]
&= \color{blue}{\frac{1}{3}}
\end{align}$$
(note that the second limit is $\color{fuchsia}{0}$ since the leading term of the sum is equivalent to $n^2$ only)
A: Another way using summation notation.
$\begin{array}\\
s(n)
&=\dfrac{\sum_{k=1}^n k(k+2)}{n^3}\\
&=\dfrac{\sum_{k=1}^n (k^2+2k)}{n^3}\\
&=\dfrac{\sum_{k=1}^n (k^2+2k+1-1)}{n^3}\\
&=\dfrac{\sum_{k=1}^n ((k+1)^2-1)}{n^3}\\
&=\dfrac{\sum_{k=1}^n (k+1)^2}{n^3}-\dfrac{\sum_{k=1}^n 1}{n^3}\\
&=\dfrac{\sum_{k=2}^{n+1} k^2}{n^3}-\dfrac{n}{n^3}\\
&=\dfrac{\sum_{k=1}^{n+1} k^2-1}{n^3}-\dfrac1{n^2}\\
&=\dfrac{\sum_{k=1}^{n+1} k^2}{n^3}-\dfrac{1}{n^3}-\dfrac1{n^2}\\
&=\dfrac{(n+1)(n+2)(2n+3)}{6n^3}-\dfrac{1}{n^3}-\dfrac1{n^2}\\
&\to \dfrac13\\
\end{array}
$
Probably too involved,
but I enjoy this kind
of playing around.
