I'm trying to find the radius of convergence of the Maclaurin series of $f(z)= \frac{2}{1+ cosh(z)} $.
I've tried substituting for z in the known series $ \frac{1}{1-z} = \sum_{n=0}^\infty{z^n}$ with known radius of convergence $|z| < 1$, and get the following:
$$ \frac{2}{1+\cosh (z)} = \sum_{n=0}^\infty{(-1)^n (e^z + e^{-z})^n}$$
$$|\cosh{z}| < 1 \Rightarrow |\frac{e^z + e^{-z}}{2}|<1 \Rightarrow |z| < \pi i$$
The radius of convergence is therefore $\pi$.
- Does this reasoning make sense?
- I'm unable to get any further with the series above. How can I convert it into a power series?
- The answer to this question states that the radius of convergence is the distance in the complex plane to the nearest singularity, which makes the problem much easier – one can just set the denominator equal to 0 and solve for $z$, which gives the same answer. However, this seems to assume that every function is analytic everywhere except at singularities, an assumption I have trouble justifying.
Best Answer
This makes little sense since: $$\cosh z\ge 1$$ along the real axis so that you cannot apply the formula for geometric series which is valid only for $|z|<1$.
The reasoning (3.) is the correct one.