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For which values of $x$ is $$ \sum_{n=1}^{\infty} \frac{(2n)!\,x^n}{2^{2n}(n!)^2}$$ convergent or divergent?

• I tried $x=-1$ and i tested it by using Comparison Test and Ratio Test but that was diverge. This series is not absolutely converge for $x=-1$ if I am right.

• Any suggestions?

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    $\begingroup$ This is the Taylor series of $\dfrac{1}{\sqrt{1-x}},$ if you start with $n=0.$ See this for more details. $\endgroup$ – Bumblebee Apr 26 '20 at 22:40
  • $\begingroup$ This could help. $\endgroup$ – Arnaud D. Apr 27 '20 at 8:08
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hint: use Rabbe's test for$ x=-1$

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If $x\ne0$, we have\begin{align}\lim_{n+\infty}\frac{\left|\frac{(2(n+1))!x^{n+1}}{2^{2(n+1)}((n+1)!)^2}\right|}{\left|\frac{(2n)!x^n}{2^{2n}(n!)^2}\right|}&=\lim_{n+\infty}\frac{(2n+2)(2n+1)}{4(n+1)^2}|x|\\&=|x|\end{align}and therefore the series converges absolutely if $|x|<1$ and diverges if $|x|>1$.

The sequence $\left(\frac{(2n)!}{2^{2n}(n!)^2}\right)_{n\in\Bbb N}$ decreases because, by the computation made above, if you divide the $(n+1)$th term by the $n$th, you get$$\frac{(2n+2)(2n+1)}{4(n+1)^2}=\frac{2n+1}{2n+2}<1.$$Also, it follows from the equality of the previous line that$$2^{2n}(n!)^2=\prod_{j=1}^n\frac{2j-1}{2j}.\tag1$$Now, note that$$\left(\frac{2j-1}{2j}\right)^2<\frac j{j+1},$$since this means that $4j^3-3j+1<4j^3$; in other words,$$\frac{2j-1}{2j}<\sqrt{\frac j{j+1}}.\tag2$$But then it follows from $(1)$ and $(2)$ that$$\frac{(2n)!}{2^{2n}(n!)^2}<\prod_{j=1}^n\sqrt{\frac j{j+1}}=\sqrt{\frac1{n+1}},$$and so, by the squeeze theorem, $\lim_{n\to\infty}\frac{(2n)!}{2^{2n}(n!)^2}=0$. Since it decreases and its limit is $0$, it follows from the Leibniz test that your series converges when $x=-1$.

But it diverges when $x=1$. You will find a proof here.

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$${\sum_{n=1}^{\infty} \dfrac{(2n)!x^n}{2^{2n}(n!)^2}} = {\frac{1+{\sqrt{1-x}}}{\sqrt{1-x}}}$$

It's will be convergent if $\vert{x}\vert < 1$, by Ratio Test

Note that this can be checked using Wolfram Alpha, with this formula: series (((2n)!x^n)/(2^(2n)(n!)^2)) from 1 to inf

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