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

taylor expansion

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?

Best Answer

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.

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