ratiotest and radius of convergence of Taylor expansion of $x\sin(x)$

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Today I helped a student who did not understand Taylor approximations. One of the exercises he had trouble with, was to determine the Taylor series of the function $$f: \mathbb{R} \to \mathbb{R}: x \mapsto x\sin(x)$$ around $0$.

The Taylor-approximation only has even powers of $x$. The solution he had, determined the radius of convergence using the ratio test. However, the ratio test uses consecutive coefficients $c_k$ (if the power series is $\sum c_k(x-a)^k$ and all $c_{2k+1}$ are zero. In the solution, the ratio of $c_{2k}/c_{2k+2}$ is computed and its limit is taken.

My question: is there a theorem which states that we can 'skip' coefficients because they are zero?

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There is the root test. Here the best thing is to note that the whole series can be written as a power series in $x^2$, and using the ratio test on this new power series yields the result for the new power series, thus for the original one.