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\title{Convergence of sinh power series}
\author{Danny Milosavljevic}
\date{\today}
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\section{Introduction}
This article shows that the convergence radius of the series A, defined as follows, is infinite for all $ z \in \mathbb{C} $:\par
\[A := \sum\limits_{n \ge 0} z^{2 n + 1}/(2 n + 1)!\]\par
\section{Proof}
\par
\[A = \sum\limits_{n \ge 0} b_n z^{2 n + 1}\]\par
with
\[b_n := \frac{1}{(2 n + 1)!}\]\par
The series A almost looks like a power series (the power series of sinh z). Compared to a normal power series, some of its coefficients are 0.\par
Hence, try to find a new function for the coefficients of the power series:\par
\[a_m = \begin{cases} \frac{1}{m!}, \text{m odd}\\ 0, \text{m even}\end{cases}\]\par
\[A = \sum\limits_{m \ge 0} a_m z^m\]\par
Using the formula of Cauchy-Hadamard, the following holds for the convergence radius R (the radius of a circle in the complex number plane):\par
\[1/R = \limsup\limits_{n \rightarrow \infty} \sqrt[n]{a_n}\]\par
\[1/R = \limsup\limits_{n \rightarrow \infty} \sqrt[n]{\begin{cases} \frac{1}{n!}, \text{n odd}\\ 0, \text{n even}\end{cases}}\]\par
I determine the accumulation point of $ \frac{1}{n!}$, let's call it h.\par
That is equal to the 1/R because it holds that\par
\[1/n! > 0\]\par
\[h = \lim\limits_{n \rightarrow \infty} \sqrt[n]{\frac{1}{n!}}\]\par
\[h = \lim\limits_{n \rightarrow \infty} \frac{1}{\sqrt[n]{n!}}\]\par
It holds that:\par
\[n! \ge \left(\frac{n}{2}\right)^{n/2}\]\par
Therefore also:\par
\[\sqrt[n]{n!} \ge \sqrt[n]{\left(\frac{n}{2}\right)^{n/2}}\]\par
Therefore also:\par
\[1/\sqrt[n]{n!} \le 1/{\sqrt[n]{\left(\frac{n}{2}\right)^{n/2}}}\]\par
\[1/\sqrt[n]{n!} \le 1/{{\left(\frac{n}{2}\right)^{1/2}}}\]\par
\[1/\sqrt[n]{n!} \le {\frac{\sqrt{2}}{\sqrt{n}}}\]\par
It holds that:
\[\lim\limits_{n \rightarrow \infty} \frac{1}{\sqrt{n}} = 0\]\par
Therefore also:\par
\[\lim\limits_{n \rightarrow \infty} \frac{\sqrt{2}}{\sqrt{n}} = 0\]\par
I resume the main calculation:\par
Because of the squeeze theorem it holds that (upper boundary first):\par
\[h \le \lim\limits_{n \rightarrow \infty} \frac{\sqrt{2}}{\sqrt{n}}\]\par
\[h \le 0\]\par
It holds that:\par
\[1/n \le \frac{1}{\sqrt{n}}\]\par
\[\lim\limits_{n \rightarrow \infty} \frac{1}{n} = 0\]\par
Because of the squeeze theorem it holds that (lower boundary):\par
\[\lim\limits_{n \rightarrow \infty} \frac{1}{n} \le h\]\par
\[0 \le h\]\par
Therefore:\par
\[h = 0\]\par
\[A = \sum\limits_{m \ge 0} a_m z^m\]\par
So the power series of A is now equal to the standard power series:\par
\[A = \sum\limits_{m \ge 0} a_m (z - z_0)^m\]\par
with\par
\[a_m = \begin{cases} \frac{1}{m!}, \text{m odd}\\ 0, \text{m even}\end{cases}\]\par
and\par\
\[z_0 = 0\]\par
The convergence radius of the circle in the complex number plane is:\par
\[1/R = h\]\par
\[1/R = 0\]\par
\[R \rightarrow \infty\]\par
(And the center is at 0).\par
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