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.

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.
