finding radius of convergence for $a_n=\frac{n^2-5}{4^n+3n}$. I am asked to find the radius of convergence for the power series
$$\sum_{n=0}^{\infty} a_nz^n$$ with $a_n=\frac{n^2-5}{4^n+3n}$.
I feel like I can use the ratio test but I am unsure because when I tried it I just got a mess.
$\frac{(n+1)^2-5}{4^{n+1}+3(n+1)}$ $\times$ $\frac{4^n+3n}{n^2-5}$
I would appreciate some help with this one.
 A: Starting where you left off:
$$\frac{(n + 1)^2 - 5}{4^{n + 1} + 3(n + 1)} \cdot \frac{4^n + 3n}{n^2 - 5}$$
$$= \frac{n^2 + 2n + 1 - 5}{4^{n + 1} + 3(n + 1)} \cdot \frac{4^n + 3n}{n^2 - 5}$$
$$=\frac{n^2 + 2n + 1 - 5}{n^2 - 5} \cdot \frac{4^n + 3n}{4^{n + 1} + 3(n + 1)}$$
Taking $n^2$ out of the first part and $4^n$ in the second part:
$$=\frac{1 + \frac{2}{n} - \frac{4}{n^2}}{1 - \frac{5}{n^2}} \cdot \frac{1 + \frac{3n}{4^n}}{4 + \frac{3(n + 1)}{4^n}}$$
So:
$$r = \lim_{n \to \infty}\left(\frac{1 + \frac{2}{n} - \frac{4}{n^2}}{1 - \frac{5}{n^2}} \cdot \frac{1 + \frac{3n}{4^n}}{4 + \frac{3(n + 1)}{4^n}}\right)^{-1}$$
Since $4^n$ increases exponentially faster than $3n$, therefore:
$$\lim_{n \to \infty}\frac{3n}{4^n} = 0$$ and $$lim_{n \to \infty}\frac{3(n + 1)}{4^{n + 1}} = 0$$
Then:
$$r = \lim_{n \to \infty}\left(\frac{1 + \frac{2}{n} - \frac{4}{n^2}}{1 - \frac{5}{n^2}} \cdot \frac{1 + \frac{3n}{4^n}}{4 + \frac{3(n + 1)}{4^n}}\right)^{-1}$$
$$= \left(\frac{1 + 0 + 0}{1 - 0} \cdot \frac{1 + 0}{4 + 0}\right)^{-1}$$
$$= \left(\frac{1}{4}\right)^{-1}$$
$$= 4$$
If I've made a mistake, can someone correct me?  I believe I haven't, but as with life, I could be wrong.  Muchly appreciated.
Edit: Thanks to @Jens's point-out,  I had forgot to do the reciprocation. (My bad)
A: By Cauchy-Hadamard,  $$r=\limsup_{n\to\infty}\dfrac 1{ {\lvert \dfrac {n^2-5}{4^n+3n}\rvert}^{1/n}}=\limsup_{n\to\infty}\dfrac 1{{(\dfrac 1{4^n}})^{1/n}}=4$$, because asymptotically $$\dfrac {n^2-5}{4^n+3n}\sim4^{-n}$$.
