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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!

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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.

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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

$$\frac12 \left (1-\frac{\log{(1+\sqrt{2})}}{\sqrt{2}} \right ) \approx 0.188387$$

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    $\begingroup$ Happy 30k, btw. $\endgroup$ – Julien Jun 17 '13 at 0:07
  • $\begingroup$ @julien: thanks! We've come a long way, haven't we? Anyway, I decided against having even one heart-attack burger to celebrate, never mind three, as I would be at risk of not seeing 40k. $\endgroup$ – Ron Gordon Jun 17 '13 at 0:12
  • $\begingroup$ Right. We need you to keep rolling and in good shape! $\endgroup$ – Julien Jun 17 '13 at 0:16
  • $\begingroup$ @julien: Shucks, that's nice. Anyway, my kids got me a hat with a turtle on it. I thought for a minute it was for reaching 30k, but then it occurred to me that today is Father's Day. $\endgroup$ – Ron Gordon Jun 17 '13 at 0:20
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    $\begingroup$ @RaymondManzoni: many thanks. Why did you refrain? This is a very nifty aspect of this particular sum and I was surprised that nobody addressed it. Best of all, in what looks like a crazy coincidence, I had just (almost) derived a closed-form expression for the series the day before, so I couldn't resist. Anyway, keep up the great work. $\endgroup$ – Ron Gordon Jun 17 '13 at 10:22
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Hint: You may use Stirling or more quickly this equivalence for central binomial coefficients as $n\to \infty$ : $$\binom{2n}{n}\sim \frac {4^n}{\sqrt{\pi n}}$$

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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.

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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.

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    $\begingroup$ No... $$\frac{4^n n!^2}{(2n)!} \sim \sqrt{\pi n}.$$ $\endgroup$ – Antonio Vargas Jun 16 '13 at 20:47
  • $\begingroup$ @AntonioVargas: What's no? $\endgroup$ – Mhenni Benghorbal Jun 16 '13 at 20:48
  • $\begingroup$ $$\lim_{n\to \infty}\frac{4^n n!^2}{(2n)!} \neq 0.$$ $\endgroup$ – Antonio Vargas Jun 16 '13 at 20:48
  • $\begingroup$ @AntonioVargas: Are you sure? $\endgroup$ – Mhenni Benghorbal Jun 16 '13 at 20:49
  • $\begingroup$ @AntonioVargas: I was considering the wrong one $$ \frac{4^n n!}{(2n)!}. $$. Thanks $\endgroup$ – Mhenni Benghorbal Jun 16 '13 at 20:51

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