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.

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 $$ \frac{a_n - a_{n-1}}{b_n - b_{n-1}} = \frac{n(n+2)}{n^3 - (n-1)^3} = \frac{n(n+2)}{n^3-n^3+3n^2 - 3n + 1} = \frac{n^2+2n}{3n^2-3n+1} \xrightarrow[n \to \infty]{}\frac{1}{3}$$
