Using Ratio test/Comparison test I have $\displaystyle\sum_{n=1}^{\infty}{\frac{(2n)!}{(4^n)(n!)^2(n^2)}}$ and need to show whether it diverges or converges. 
I attempted to use the ratio test, but derived that the limit of $\dfrac{a_{n+1}}{a_n}=1$ and hence the test is inconclusive. 
So I now must attempt to use the comparison test, but I am struggling to find bounds to compare it to to show either divergence or convergence. 

See below for my solution. Definition:
$$\lim_{x\rightarrow\infty} \frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty} \frac{\frac{(2n+2)!}{4^{n+1}\cdot((n+1)!)^2 \cdot (n+1)^2}}{\frac{(2n)!}{4^n\cdot (n!)^2 \cdot n^2}}$$
Expand the factorials, to prepare for removal due to ratio.
$$=\lim_{n\rightarrow\infty} \frac{\frac{(2n+2)(2n+1)\cdot(2n)!}{4\cdot 4^n\cdot (n+1)^2 \cdot (n!)^2 \cdot (n+1)^2}}{\frac{(2n)!}{4^n\cdot (n!)^2 \cdot n^2}}$$
Cancel out the factorial operations:
$$=\lim_{n\rightarrow\infty} \frac{\frac{(2n+2)(2n+1)}{4\cdot (n+1)^2 \cdot (n+1)^2}}{\frac{1}{n^2}}$$
Put into simple ratio form for final simplification:
$$=\lim_{n\rightarrow\infty} \frac{(2n+2)(2n+1)\cdot n^2}{4\cdot (n+1)^4}$$
Expand the powers of polynomials:
$$=\lim_{n\rightarrow\infty} \frac{4n^4+6n^3+2n^2}{4\cdot (n^4+4n^3+6n^2+4n+1)}$$
Approximation for simplicity. Throwing out the lower order positive terms in the denominator makes the fraction larger, so implied comparison test. Ditto for swallowing the numerator terms in the highest order.
$$\approx\lim_{n\rightarrow\infty} \frac{4n^4}{4n^4}\rightarrow 1$$
Therefore, the test is inconclusive.
