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.)
Best Answer
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 $n \to \infty$,
as announced.