Showing that the radius of convergence of a power series is $R=\infty$

Question

Let $a_k,z\in \mathbb{C}$ and $A(z) = \sum_{k=1}^{\infty}a_kz^k$ with radius of convergence $R$.

For each $x \in \mathbb{R}$ with $0 \le x < R$, $A(x) = e^x > -2e^{-2x}$.

1. prove $R=\infty$

2. find $A(i\pi)$

My attempt was to use $A(x)$ with Taylor expansion, but it didn't really help me much. How should I attack this problem?

Answer 1

You can use Taylor series of exp : $$e^z=\sum_{k=0}^\infty \frac{z^n}{n!}$$ Next step you calculate $A(x)$ using this series and you will have what you want.

Answer 2

It is well known that the power series for $e^z=\sum z^n/n!$ converges everywhere. For the computation, use the Hadamard formula: $r=1/\limsup1/\sqrt[n]{n!}=1/1/\infty=1/0=\infty$.

By the identity theorem, $A(z)=e^z-2e^{-2z}$.

It follows that $A(i\pi)=e^{i\pi}-2e^{-2i\pi}=-1-2=-3$.

Answer 3

By the assumption, $A(z)$ is an analytic function in the circle $|z|<R$. Along the line segment connecting $0$ to $R$, it coincides with $e^x-2e^{-2x}$. The latter function is an entire function, in particular it is analytic in $|z|<R$. Thus, by the uniqueness properties of analytic functions, $A(z)=e^z-2e^{-2z}$ on $|z|<R$. But since $e^z-2e^{-2z}$ is entire, its Taylor series at the origin converges for all $|z|$. The original power series defining $A(z)$ must be the Taylor series of the function at the origin, so its radius of convergence $R$ is infinity. Also $A(\pi i) = e^{\pi i} - 2e^{ - 2\pi i} = - 1 - 2 = - 3$.