simplifying series using factorials to see if it is converges conditionally or absolutely or diverges? $$\sum_{n=0}^\infty {4^n(n!)^2\over (2n)!}$$
Would I be able to cross out one of the n factorials so the numerator would have ${n!}$ instead of ${(n!)^2}$. I haven't learned much about factorials so this is pretty new to me. 
 A: Let the $n$-th term be $a_n$. Note that 
$$\frac{1}{a_n}=\frac{(2n)!}{4^n n!n!}=\binom{2n}{n}\frac{1}{2^{2n}}.$$
We recognize the last expression as the probability of getting $n$ heads and $n$ tails when we toss a fair coin $2n$ times.
In particular, $\frac{1}{a_n}\lt 1$ if $n\gt 0$, and therefore $a_n\gt 1$.
The terms $a_n$ do not have limit $0$, so our series diverges.
A: No.  I suggest you write out the term for $n=3$ to see why.
A: The way to go here is the ratio test; trying to "guess" convergence based on how the summand looks is not a good idea. The ratio test will tell you whether a series (absolutely) converges or not. What ratio test says is that if $\lim_n \left|\dfrac{a_{n+1}}{a_n}\right| < 1$, the series converges absolutely (where the $a_n$ are the terms in the series).
In your case, you want to inspect the nature of
$$\lim_{n\to\infty}\frac{4^{n+1}((n+1)!)^2}{(2(n+1))!}\frac{(2n)!}{4^n(n!)^2}.$$
If you make use of factorial properties, you will get your answer.
A: No. Imagine that you are multiplying all the numbers from 2n until n. From then on the ratio will be one. What you are left with is the product of the numbers (2n)(2n-1)(2n-2)...(n+1), which is obviously larger than (n)(n-1)...1
A: You can, but it doesn’t really help much: the denominator is then the rather ugly
$$(2n)(2n-1)\ldots(n+2)(n+1\;.$$ 
Notice that $4^n=2^n\cdot2^n$ and
$$2^nn!=(2n)(2n-2)(2n-4)\ldots(4)(2)\;,$$
so the $n$-th term can be written
$$\frac{(2n)(2n-2)\ldots(4)(2)}{(2n-1)(2n-3)\ldots(3)(1)}=\frac{2n}{2n-1}\cdot\frac{2n-2}{2n-3}\cdot\ldots\cdot\frac12\;.$$
A: A small trick which is (at least to me) very useful when handling expressions which involve factorials is Stirling simplest approximation which write $$m!=\sqrt{2 \pi } e^{-m} m^{m+\frac{1}{2}}$$ Applied to the expression $$a_n={4^n(n!)^2\over (2n)!}\approx \sqrt{\pi } \sqrt{n}$$ and so $$\sum_{n=0}^p a_n\approx \sqrt{\pi } H_p^{\left(-\frac{1}{2}\right)}$$ where appears the harmonic number.
For $p=1000$, the exact partial sum gives $37408.0$ while the approximation leads to $37394.3$.
