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Consider the power series $\sum a_nx^n $ where $$a_n = \frac{2.4.6...(2n)}{1.3.5...(2n+1)}$$ Now the ratio test shows absolute convergence for all $|x|<1$ and divergence for $|x|>1$. Also Raabe’s test shows divergence for x=1. The remaining case is for x=-1 which I’m not able to solve. Help will be highly appreciated.(using only basic inequalities/methods. No sterling’s formula etc :) )

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  • $\begingroup$ It can be shown that $\frac12\sqrt{\frac\pi{n+1}}\le a_n\le\frac12\sqrt{\frac\pi{n+1/2}}$ $\endgroup$
    – robjohn
    Commented Oct 2, 2018 at 5:12

2 Answers 2

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The series is alternative, so we may consider the Leibniz test. Pretty clear that the series converges for $x =-1$. Could you find out why?

UPDATE

To show $a_n\to 0$, one may try prove that [WRONG INEQUALITY] $$ \frac {(2n)!!}{(2n+1)!!} < \frac 1{\sqrt {2n+1}}. $$

Also the series diverges at $x = 1$ as the OP mentioned by Raabe test.

EDIT

Thanks to @robjohn who pointed out the inequality is reversed. It should be $$ \frac {(2n)!!}{(2n+1)!!} < \frac 1{\sqrt {n+1}}. $$

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  • $\begingroup$ I’m not able to find a bound for a_n $\endgroup$ Commented Oct 2, 2018 at 4:12
  • $\begingroup$ +1 Alternatively, the series is absolutely convergent, as the $x = 1$ case indicates. $\endgroup$ Commented Oct 2, 2018 at 4:16
  • $\begingroup$ How will you show an goes to 0? $\endgroup$ Commented Oct 2, 2018 at 4:30
  • $\begingroup$ @JohnMitchell Sorry about my wrong comment. $\endgroup$
    – xbh
    Commented Oct 2, 2018 at 4:31
  • $\begingroup$ an is certainly decreasing but that does not show it goes to 0. $\endgroup$ Commented Oct 2, 2018 at 4:31
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Using Simple Estimates

Let $$ a_n=\frac1{3/2}\frac2{5/2}\cdots\frac{n}{n+1/2} $$ and $$ b_n=\frac{3/2}2\frac{5/2}3\cdots\frac{n+1/2}{n+1} $$ and $$ c_n=\frac{1/2}1\frac{3/2}2\cdots\frac{n-1/2}{n} $$ Since $\frac{x}{x+1/2}\lt\frac{x+1/2}{x+1}$, we get $$ c_n\lt a_n\lt b_n $$ Therefore, $$ \frac1{2n+1}=a_nc_n\le a_n^2\le a_nb_n=\frac1{n+1} $$ This means that $$ \sum_{n=0}^\infty a_nx^n $$ converges for $-1\le x\lt1$.


Applying More Power

Since $$ a_n=\frac{\Gamma\!\left(\frac32\right)\Gamma(n+1)}{\Gamma(1)\,\Gamma\!\left(n+\frac32\right)} $$ and by the log-convexity of $\Gamma$, we get $$ \frac12\sqrt{\frac\pi{n+1}}\le a_n\le\frac12\sqrt{\frac\pi{n+1/2}} $$

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