Suppose the power series $ y=\sum^{\infty}_0 a_nx^n\;$ has terms that satisfies the recurrence relation: $$a_{n+2}=\frac{(2n_1)(3n+2)}{(n+3)(2n-5)}a_n$$ with $a_0=1$,$a_1=0$. What is the radius of convergence $R$ of the power series?


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    $\begingroup$ what is $n_1$? is it $a_{n+1}$? $\endgroup$ – windircurse Jul 17 '16 at 15:08

Since $a_{n+2}=\frac{2n(3n+2)}{(n+3)(2n-5)}\,a_n$, with $a_1=0$, we find that all of the odd terms in the series are zero.

Let $b_n=a_{2n}$ so that $y=\sum_{n=0}^\infty b_nx^{2n}$. Then, $b_{n+1}=\frac{4n(6n+2)}{(2n+3)(4n-5)}b_n$ and the ratio test gives

$$\lim_{n\to \infty }\left(\frac{b_{n+1}x^{2n+2}}{b_nx^{2n}}\right)=3x^2<1$$

whenever $|x|<1\sqrt{3}$.