finding the radius of convergence of a power series

I have been trying to find out the radius of convergence ($R$) of the power series $\sum a_{n}x^{n}$, where $a_{0}=0$, $a_{n}=\sin(n!)/n!$ for $n\geq 1$. I am aware of the formula for finding $R=\limsup a_{n}/a_{n+1}$. I am also aware of Cauchy-Hadamard formula. But I could not evaluate the value. Please help.

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Your formula can't be right. For one thing, it predicts that every power series with alternating zero and non-zero terms has an infinite radius of convergence! Your formula is only valid if $(a_n)$ is monotonic (which is not the case here). – TonyK Nov 25 '12 at 6:57

For the power series, $$f(z) = \sum_{k=0}^{\infty} a_k z^k$$ the radius of convergence is given by $R$ such that $$\limsup_{n \to \infty} \left \vert a_n \right \vert^{1/n} R = 1$$ i.e. $$R = \dfrac1{\limsup_{n \to \infty} \left \vert a_n \right \vert^{1/n}}$$ We have $a_n = \dfrac{\sin(n!)}{n!}$. Note that since $\vert \sin(x) \vert \leq 1$, we get that $\vert a_n \vert \leq \dfrac1{n!}$. Hence, $$\vert a_n \vert^{1/n} \leq \dfrac1{(n!)^{1/n}}$$ Hence, $$0 \leq \limsup_{n \to \infty} \left \vert a_n \right \vert^{1/n} \leq \limsup_{n \to \infty} \dfrac1{(n!)^{1/n}} = 0$$ We used the fact that $\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite to get the last step.
Hence, the radius of convergence is $\infty$.