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. 
 A: The expression $\frac{(2n)!}{4^n(n!)^2}$, rewritten as $\binom{2n}{n}\frac{1}{2^{2n}}$, can be recognized as the probability of an equal number of heads and tails in $2n$ tosses of a fair coin. In particular it is less than $1$.
Thus the $n$-th term $a_n$ of the sequence we were given is less than $\frac{1}{n^2}$. Now use the Comparison Test.
A: Hint: Let $a_n=\dfrac{(2n)!}{(4^n)(n!)^2(n^2)}$ then the sequence $(n^2a_n)$ is decreasing (can you show this?) hence $a_n\leqslant\dfrac{a_1}{n^2}$ for every $n$ and...
A: We'll use the limit criteria:
"Let $\sum a_n$ and $\sum b_n$ be series of positive terms. Suppose 
$$\lim_{n\to \infty} \frac{a_n}{b_n}=l\in\mathbb R^+ \cup \{+\infty\}.$$
Then:


*

*If $l\in\mathbb R$ and $l>0$, then both series have the same behavior.

*If $l=0$, the convergence of $\sum b_n$ implies the convergence of $\sum a_n$.

*If $l=+\infty$, the convergence of $\sum a_n$ implies the convergence of $\sum b_n$."


For this case, let $$a_n = \frac{(2n)!}{(4^n)(n!)^2(n^2)}, b_n=\frac {1}{n^2}$$
Then $$\lim_{n\to \infty} \frac{a_n}{b_n}=\lim_{n\to \infty} \frac{(2n)!}{4^n (n!)^2}=0$$
By the limit criteria, the convergence of $\sum \frac{1}{n^2}$ implies the convergence of $\sum \frac{(2n)!}{(4^n)(n!)^2(n^2)}$.
A: I can provide a different perspective that may or may not answer your question, but does give you a place to start.
Notice that 
$$
\frac{(2n)!}{4^n n!^2 n^2} = \frac{{2n \choose n}}{4^n n^2}
$$
So 
$$
\sum_{n=1}^\infty \frac{{2n \choose n}}{4^n n^2} 
$$
Note that
$$
\frac{{2n+2 \choose n+1}}{{2n \choose n}} = \frac{2(\pi n^4 - \pi n^2 - \sin(pi n))\Gamma(n-1)}{\sqrt{\pi}(n+1)\Gamma(n+\frac{1}{2})}
$$
With the ratio test
$$
\lim_{n \rightarrow \infty} \frac{8(n+1)\Gamma(n-1) 
(\pi n^4 - \pi n^2 - \sin(\pi n))}{\sqrt{\pi} \times 2^{2n} \Gamma(n+\frac{1}{2})4^n n^2}
$$
$$
\lim_{n \rightarrow \infty} \frac{8(n+1)(n-2)!(\pi n^4 - \pi n^2 - \sin(\pi n))}{\sqrt{\pi}\Gamma(n+\frac{1}{2}) 16^n}
$$
$$
\lim_{n \rightarrow \infty} \frac{4(n+1)(2n+1)(n-2)!(\pi n^4 - \pi n^2 - \sin(\pi n))}{\sqrt{\pi}(\frac{1}{2}(2n+1))! 16^n}
$$
If you were to plot this on wolframalpha, this appears to approach zero. Therefore the series would converge, if you would like more explicit proof you could attempt to solve the limit, i would but it would probably take a while.... 
A: 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.
