# Power series convergence (hypergeometric)

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 :) )

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

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 $$\frac {(2n)!!}{(2n+1)!!} < \frac 1{\sqrt {n+1}}.$$

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

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 $$\frac12\sqrt{\frac\pi{n+1}}\le a_n\le\frac12\sqrt{\frac\pi{n+1/2}}$$