Finding a radius of convergence of power series I have  to find the radius of convergence of some power series but I find myself in trouble for three of them : the series are 


*

*$\sum2^kx^{k!}$

*$\sum\sinh(k)x^k$ 

*$\sum\sin(k)x^k$.


For the first one I have tried the Ratio Test for series first but I don't kow how to deal with the factorials as exponents... For the second one, I'm tempted to say it never converges but I cannot prove it. And for the last one, I'd say it converges for $|x|< 1$ since $-1<\sin(h)<1$ anyway, but I'm not quite sure...
Some hints would help a lot! Thank you.
 A: The first one is a lacunary series, it has a lot of zero coefficients. For such series, the ratio test cannot work. The generally applicable Cauchy-Hadamard formula, however, has no problems with the first series. The Cauchy-Hadamard formula says
$$\frac{1}{R} = \limsup_{n\to\infty} \lvert a_n\rvert^{1/n},$$
where $R$ is the radius of convergence, and the extreme cases are to be interpreted $\frac{1}{0} = \infty$ resp. $\frac{1}{\infty} = 0$. Here, the nonzero coefficients occur at the factorial indices, and
$$\limsup_{k\to\infty} \left(2^k\right)^{1/k!} = \limsup_{k\to\infty} 2^{1/(k-1)!} = \;?$$
For the second one, we can also use the Cauchy-Hadamard formula, or we can use the ration test, whichever way one chooses, the essential fact is that
$$\sinh t = \frac{e^t-e^{-t}}{2}$$
behaves for large $t$ essentially as $e^t$ ($e^t/2$, if one needs more accuracy).
For the third, the ratio test is not adequate, since the sine oscillates between $-1$ and $1$, but the Cauchy-Hadamard formula quickly settles the matter.
