Radius of convergence in a series. Ratio test. I am having a hard time with this question. 
$$\sum_{k=0}^{\infty} \frac{-(1)^k (4^k -3)x^{2k}}{k^4+3}$$
I used the ratio test and got stuck here:
$$x^2 \lim_{k\to\infty} \frac {(4^{k+1}-3)(k^4+3)}{((k+1)^4+3)(4^k-3)}$$ 
Im not sure how to simplify it further.

$$\lim_{k\to\infty} \frac {(4^{k+1}-3)(k^4+3)}{((k+1)^4+3)(4^k-3)}=\lim_{k\to\infty} \frac {(4^{k+1}-3)}{(4^k-3)}\cdot \lim_{k\to\infty} \frac {(k^4+3)}{((k+1)^4+3)}$$
as long as the two limits on the right hand side exist. But,
 $$\lim_{k\to\infty} \frac {(4^{k+1}-3)}{(4^k-3)}=\lim_{k\to\infty} \frac {(4-3/4^k)}{(1-3/4^k)}=4$$
and
$$\lim_{k\to\infty} \frac {(k^4+3)}{((k+1)^4+3)}=\lim_{k\to\infty} \frac {(1+3/k^4)}{((1+1/k)^4+3/k^4)}=1.$$
