What is the radius of the power series $\sum_{n=0}^\infty\frac{\tanh^{(n)}(0)}{n!} z^n$? Justify your answer.
My steps toward a solution
I can express $\tanh$ simpler as:
\begin{align*}
\tanh z
&=\frac{\sinh z}{\cosh z} \\
&=\frac{e^z-e^{-z}}{e^z+e^{-z}}
\end{align*}
However I'm not sure what the $n^{th}$ derivative of this would be.
I also know Cauchy's Formula, which says for a Jordan curve $J$ and holomorphic function $f$ on $J \cup Int(J)$, we can compute for any $a \in Int(J)$
\begin{equation}
f(a)=\frac{1}{2\pi i} \oint_J \frac{f(z)}{z-a}\, dz
\end{equation}
but I'm not sure how to apply that.
I also have some other formulas,
\begin{equation}
f(a)=\sum_{n=0}^\infty \left(a-a_0\right)^n \frac{1}{2 \pi i} \oint_J \frac{f(z)}{\left(z-a_0\right)^{n+1}}\, dz
\end{equation}
for $|a-a_0|< dist(a_0,J)$
and
\begin{equation}
f^{(n)}\left(a_0\right)=\frac{n!}{2 \pi i} \oint_J \frac{f(z)}{\left(z-a_0\right)^{n+1}}\, dz
\end{equation}
for $a_0 \in Int(J)$.
However, I'm not sure how I would apply these formulas either (if that's actually what I'm supposed to do!). Thanks for helping me understand this and work this out.
Best Answer
The radius of convergence is the distance of $0$ to the closest singularity of $\tanh z$. The singularities of $\tanh z$ are the zeroes of $\cosh z$. These are $$ \pm\frac{\pi}{2}\,i+2\,k\,\pi\,i,\quad k\in\mathbb{Z}. $$ The radius of convergence is thus $\pi/2$.