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I'm trying to analyze the ratio of convergence of the following power series, I also want to know if the series converges at the endpoints of R (radius of convergence).

$$ \sum_{n=0}^ \infty \frac{x^n (2n)!}{(n!)^2}$$

By the ratio test, it is easy to see that$\ R=1/4$, that is, the series converges for$\ |x|<1/4$

I want to know what happens with the series at$\ x=1/4$ and$\ x=-1/4$

At$\ x=1/4$

$$ \sum_{n=0}^ \infty \frac{(1/4)^n (2n)!}{(n!)^2}$$

Do I have to use the ratio test yet again?

DavidTomahawk
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    I do not recommend using the ratio test again because it will give the limit of ratio $1$. – Sungjin Kim Jun 13 '18 at 01:58
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    It's called a radius of convergence :-) – N8tron Jun 13 '18 at 01:59
  • Divergence test may be useful. – AnyAD Jun 13 '18 at 02:00
  • What would you recomend @i707107 ? – DavidTomahawk Jun 13 '18 at 02:00
  • I recommend using $(2n)!= (2n\cdot ( 2n-2 ) \cdots 6 \cdot 4 \cdot 2) \cdot ((2n-1)\cdot (2n-3) \cdots 5 \cdot 3 \cdot 1)$, and $2^n n! = 2n\cdot ( 2n-2 ) \cdots 4 \cdot 2$. – Sungjin Kim Jun 13 '18 at 02:03
  • Hints: $\enspace (1)\enspace $$\sum\limits_n\left(\frac{(1/4)^n (2n)!}{(n!)^2}-\frac{1}{\sqrt{\pi n}}\right)$ is convergent and therefore $\sum\limits_n (\frac{(1/4)^n (2n)!}{(n!)^2}$ divergent. $\enspace (2)\enspace $ $ \sum\limits_n\frac{(-1/4)^n (2n)!}{(n!)^2}$ is convergent by Leibniz criteria (= alternating series test). – user90369 Jun 13 '18 at 11:04

2 Answers2

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with using stirling approximation $$\sum_{n=0} \frac{(1/2)^{2n}(2\pi2n)^{1/2}(2n)^{2n}e^{-2n}}{2\pi n (n)^{2n}e^{-2n}} = \sum_{n=0}1/(\pi n )^{1/2}$$ diverges because p series (p<1). However, I am not sure if the my answer is true or not. edit: Besides, for the x=-1/4, alternating series converges. Because its limit at infinity is 0 and monotonic decreasing sequence.

superabsorbent polymer
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We have $$ \int_{0}^{\pi/2}\left(\cos x\right)^{2n}\,dx = \frac{\pi}{2}\cdot \frac{\binom{2n}{n}}{4^n}\sim \frac{C}{\sqrt{n}} \tag{1}$$ from which it follows that $$ \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}z^n = \frac{1}{\sqrt{1-z}}\tag{2} $$ for any $z\in(-1,1)$. The LHS of $(2)$ is convergent at $z=-1$ by Leibniz' test and divergent at $z=1$.

Jack D'Aurizio
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