Testing convergence of $\sum_{n=0}^{\infty }(-1)^n\ \frac{4^{n}(n!)^{2}}{(2n)!}$ Can anyone help me to prove whether this series is convergent or divergent:
$$\sum_{n=0}^{\infty }(-1)^n\ \frac{4^{n}(n!)^{2}}{(2n)!}$$
I tried using the ratio test, but the limit of the ratio in this case is equal to 1 which is inconclusive in this case. Any hints please!
 A:  Hint: Use the alternating series convergence test.
The work you already did is not wasted: $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{4(n+1)^2}{(2n+2)(2n+1)}=\frac{2n+2}{2n+1}>1$$
Hence the terms do not approach 0 in absolute value, and by the $n^\textrm{th}$ term test, the series diverges.
A: I understand why the series was demonstrated to diverge.  But please allow me to be devil's advocate here and demonstrate that the sum may be assigned a finite value in the sense of an analytic continuation.  
I refer to the result I derived yesterday in connection with this problem:
$$\frac{x \, \arcsin{x}}{(1-x^2)^{3/2}} + \frac{1}{1-x^2} = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} x^{2 n}$$
Then making the substitution $x \mapsto i x$ I get
$$\sum_{n=0}^{\infty} (-1)^n \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} x^{2 n}  = \frac{1}{1+x^2} - \frac{x \, \log{(x+\sqrt{1+x^2})}}{(1+x^2)^{3/2}}$$
The radius of convergence of this series is in fact $1$.  Well, sort of.  Obviously, at $x=1$, the series in fact diverges as demonstrated above.  But the limit of the sum as $x \to 1^-$ exists and is equal to 
$$\frac12 \left (1-\frac{\log{(1+\sqrt{2})}}{\sqrt{2}} \right ) \approx 0.188387$$
A: Hint: You may use Stirling or more quickly this equivalence for central binomial coefficients as $n\to \infty$ :
$$\binom{2n}{n}\sim \frac {4^n}{\sqrt{\pi n}}$$
A: By Stirling's approximation $n!\sim\sqrt{2\pi n}(n/e)^n$, so 
$$\frac{4^{n}(n!)^{2}}{(2n)!}\sim \frac{2\pi n 4^{n} (n/e)^{2n}}{\sqrt{4\pi n}(2n/e)^{2n}} =\sqrt{\pi n}.$$ Thus, the series diverges.
A: Hint: You cannot use the alternating series test. Since 
$$ \lim_{n\to \infty}\frac{4^n n!^2}{(2n)!} \neq 0. $$ You can use the identity
$$ (2n)!=\Gamma(2n+1)={\frac {n  {4}^{n}\Gamma  \left( n \right) \Gamma 
 \left( \frac{1}{2}+n \right) }{\sqrt {\pi }}},$$
or Stirling approximation to see this.
