Radius of convergence for recursively defined power series

Question

Given $\frac{a_{n+2}}{a_n}=\frac{n(n-1)+1}{(n+1)(n+2)}$, how may one find the radius of convergence for $\sum_{n=0}^{\infty} a_n x^n$? The problem would of course be easy if the numerator was $n+1$ rather than $n+2$, but alas! My lecturer used something to do with splitting the series into its even part and its odd part... But I really wasn't able to follow what was being done.

Answer 1

Let $r_n=\frac{a_{n+1}}{a_{n}}$, then $$r_nr_{n+1}=\frac{a_{n+2}}{a_n}=\frac{n(n-1)+1}{(n+1)(n+2)}.$$ Assuming that the series has a radius of convergence $R$ and $\frac{1}{R}=\lim_{n \to \infty}r_n$. In which case, we have $$\frac{1}{R^2}=\lim_{n \to \infty}r_nr_{n+1}=\lim_{n \to \infty}\frac{n(n-1)+1}{(n+1)(n+2)}=1.$$ So $R=1$.

Answer 2

The given recurrence is precisely the expression of the ratio test. As the limit of the fraction is $1$, so is the radius of convergence.

Now the recurrence only relates terms of the same parity so that in fact you have two interleaved sequences ($a_0,a_1$ can be specified independently). These two sequences obviously have the same radius and so has their sum.