Radius of Convergence, putting equation into the right order 
Problem:
$\sum\limits_{k=0}^\infty(k+2)^2\cdot\frac{(3x+1)^k}{2^k\cdot2^2}$
Determine whether this is convergent or divergent. Use the Radius of Convergence

Hey there,
I ran into this Exercise and found myself a little bit of trouble.
According to my teacher, before I can use the Radius of Convergence, I need to put this equation into a form like:

$\sum\limits_{k=0}^\infty a_n\cdot x^k$

I'm a bit embarassed but I can't transform this. I just follow up with my thoughts and how far I went.
$\sum\limits_{k=0}^\infty(k+2)^2\cdot\frac{(3x+1)^k}{2^k\cdot2^2}$
= $\sum\limits_{k=0}^\infty(k^2+4k+4)\cdot\frac{(3x+1)^k}{2^k\cdot4}$
= $\sum\limits_{k=0}^\infty\frac{(k+2)^2}{4}\cdot\frac{(3x+1)^k}{2^k}$
Now I would do $\sqrt[k]{\frac{(3x+1)^k}{2^k}}$ But that would eliminate my $x^k$
Kind of running circles here.
Any help is appreciated
 A: You’re taking your teacher’s statement a little too literally. What it really means is that you need to get it in the form $\sum\limits_{k\ge 0}a_ky^k$, where the coefficients $a_k$ don’t depend on $x$, and $y$ does. In this problem you want to rewrite it as follows:
$$\sum_{k=0}^\infty(k+2)^2\cdot\frac{(3x+1)^k}{2^k\cdot2^2}\sum_{k\ge 0}\left(\frac{k+2}2\right)^2\left(\frac{3x+1}2\right)^k\;.$$
Here $a_k=\left(\dfrac{k+2}2\right)^2$ and $y=\dfrac{3x+1}2$. When you apply the ratio test, your $(k+1)$-st term will be
$$\left(\frac{(k+1)+2}2\right)^2\left(\frac{3x+1}2\right)^{k+1}\;.$$
You’ll end up getting a bound on $|y|$, i.e., on $\left|\dfrac{3x+1}2\right|$, something like $$\left|\frac{3x+1}2\right|<c\;;$$ solve that inequality for $x$, and you’ll get something of the form $a<x<b$, so that the radius of convergence is $\dfrac{b-a}2$, half the length of the interval of convergence.
A: First recall that
$$\sum_{n=1}^{\infty} a_n$$ converges if $$\limsup_{n} \vert a_n \vert^{1/n} < 1$$
In your case, your $a_n = (n+2)^2 \dfrac{(3x+1)^n}{2^n \cdot 2^2}$.
Note that $$\left \vert \dfrac{(n+2)^2}{2^2}\right \vert^{1/n} \to 1 \,\,\,\,\,\,\,\,(\text{Why?})$$
Hence, $$\left \vert (n+2)^2 \dfrac{(3x+1)^n}{2^n \cdot 2^2}\right \vert^{1/n} < 1 \implies \left \vert \dfrac{3x+1}2\right \vert < 1 \implies -2 < 3x+1 < 2$$
Hence,
$$-1 < x < \dfrac13$$ is your desired interval of convergence.
