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The problem is, let $\{a_n\}$ be defined recursively with $a_0 = a_1 = 1$ and

$$a_{n+2} = \frac{2n-1}{(n+1)(n+2)} a_n$$

Show $\sum_{n \geq 0} a_nx^n$ converges for all $x$. The following is my approach.

Clearly $$\lim \left|\frac{a_{n+2}}{a_n}\right| = 0$$

and so by the ratio test for any arbitrary but fixed non-zero $x \in \mathbb{R}$

$$\sum_{n \ \text{odd}}^{} a_nx^n \ \text{and} \ \sum_{n \ \text{even}}^{} a_nx^n$$

both converge (the zero case is trivial). Hence we can simply add up the two series to see $\sum_{n \geq 0} a_nx^n$ converges.

My question is, is this formally valid? We're rearranging an infinite series.

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    if the convergences is uniform everything is fine – 2017-02-13
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    Yes, it's valid. You've shown that both parts are absolutely convergent for all $x$, hence the full thing is absolutely convergent, and may be rearranged ad libitum. – 2017-02-13

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It's valid because you can use the same argument to show that the sums of odd and even terms are uniformly convergent (i.e. $\sum_{\text{odd }n}|a_nx^n|$ is convergent). Then it follows that the whole series is absolutely convergent since $\sum_{n=1}^N|a_nx^n|=\sum_{\text{odd }n\leq N}|a_nx^n|+\sum_{\text{even }n\leq N}|a_nx^n|$, which is bounded by the sum of the values of the infinite odd and even sums.

Once you know a series is uniformly convergent you can reorder the terms without changing the sum. (This doesn't work in general, e.g. the terms of $\sum \frac{(-1)^n}n$ can be rearranged to give any sum you want. That series is not absolutely convergent because $\sum\frac 1n$ diverges.)