Finding the limit of the ratio of a sequence. I don't have any experience with formal mathematics (I took calculus in college, but that's about it), and I've come across a problem in my job that I don't quite know how to tackle.
I have a sequence defined as follows:
$$a_{n+1} = 2a_n + n^2$$
$$a_1 = 12$$
And I'm looking for the limit of the ratio of consecutive terms in this sequence. The sequence goes as follows:
$$12, 25, 54, 117, 250, 525, \ldots$$
With the ratio of consecutive terms going as follows:
$$2.08, 2.16, 2.17, 2.14, 2.10, 2.07, \ldots$$
I wrote the sequence out to many terms in Excel and it looks like the ratio goes to exactly $2$. My supervisor is kind of convinced that this is true, but I think both she and I would feel more comfortable with a mathematical explanation.
I tried manipulating the sequence as follows:
$$\frac{a_{n+1}}{a_n} = 2 + \frac{n^2}{a_n}$$
It's clear that as $n \to \infty$, the value of $2 \to 2$. But why does the value of $\frac{n^2}{a_n} \to 0$?
Can someone explain this in a way that's mathematically tractable to someone like myself who barely knows any mathematics? Anything up until calculus should be okay, but I don't know all the new things kids are learning these days (see: set theory).

(Sidenote: I've seen a few posts here in passing and I'm in love with the Stack Exchange network, and I'm so happy you guys use $\LaTeX$ style expressions here. I wasn't very math-inclined in college, but I had a part-time job transcribing documents into $\LaTeX$ for math and physics nerds majors.)
 A: Hint. From the identity,
$$
a_{n+1} = 2a_n + n^2, \qquad n=1,2,3,\ldots, \tag1
$$ one may obtain
$$
\frac{a_{n+1}}{2^{n+1}}- \frac{a_n}{2^n}= \frac{n^2}{2^{n+1}}, \tag2
$$ then, by telescoping,
$$
\frac{a_n}{2^n}- \frac{a_1}{2}=\sum_{k=1}^{n-1}\frac{k^2}{2^{k+1}}, \tag3
$$ then, by using the evaluation here, one has
$$
\frac{a_n}{2^n}-6=3-\frac3{2^n}-\frac{2n}{2^n}-\frac{n^2}{2^n} \tag4
$$ giving, as $n \to \infty$,

$$
\frac{a_{n+1}}{a_n}=\color{red}{2}+\frac{n^2}{9 \times 2^n-n^2-2 n-3} \longrightarrow \color{red}{2} \tag5
$$ 

as announced.
A: The difference equation $$a_{n+1}-2a_n=n^2\tag1$$ can be solved by first finding the general solution $a_n=k2^n$ for the homogeneous equation
$$
a_{n+1}-2a_n=0,\tag2$$
where $k$ is a free parameter to be determined later, and then guessing a particular solution of the form $A_n=cn^2+dn+e$ to the inhomogeneous equation (1). Plugging this form for $A_n$ into (1) and equating coefficients gives us three equations in the three unknowns $c,d,e$. Solving, we get $c=-1$, $d=-2$, and $e=-3$, so $A_n=-(n^2+2n+3)$. 
The solution to the original equation is then the sum of the general and particular solutions:
$$
a_n = k2^n-(n^2+2n+3)\tag3
$$
With the initial condition $a_1=12$ we find $k=9$. Since $a_n$ grows exponentially this explains why $n^2/a_n\to0$ as $n\to\infty$.
A: Other answers are good enough, but here is another solution which may be easier to follow.
By definition:
$$ \begin{align}
 a_{n+1} &= 2a_n+n^2 \\ &> 2a_n > 2^2a_{n-1} > \dots  > 2^na_1
\end{align} $$
Therefore:
$$ 0<\frac{n^2}{a_n}<\frac{2}{a_1}\frac{n^2}{2^n}$$
Since $2/a_1$ is constant and $n^2/2^n\to0$ as $n\to\infty$, by the squeeze theorem, $\lim_{n\to\infty} n^2/a_n=0$ as well.
