# 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.

• Thanks for the edit user127001, it makes it much clearer but I didn't have LaTeX available here myself. – Kika Jan 7 '14 at 14:20

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.

• I have no notion of lim sup and I'm not sure I get what you're doing with the first series... Can you explain? – Kika Jan 7 '14 at 14:58
• In the first series, I'm ignoring all zero coefficients, since $\lvert a_n\rvert^{1/n}=0$ if $a_n=0$, only the nonzero coefficients influence the radius of convergence. The limes superior of a sequence of real numbers is the largest accumulation point of the sequence, formally, $$\limsup_{n\to\infty}b_n=\lim_{n\to\infty}\left(\sup_{k\geqslant n}b_k\right).$$ It's an important concept, and I'm surprised it hasn't been treated at a point where you're asked to determine the radius of convergence of a power series. Read up on it in a good book. You will need it often. – Daniel Fischer Jan 7 '14 at 15:11