Laurent expansion for $1/\cos(z)$ I have a quick question.
How to find the Laurent expansion for $1/\cos(z)$
In the link above the person asks how to find the Laurent expansion for $\frac{1}{cos(z)}$.
The accepted answer utilizes the fact that \begin{align}
\frac{1}{\sin t}&=\frac{1}{ t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\dots } \\
                &=\frac{1}{t}\frac{1}{1-\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)}\\
                &=\frac{1}{t}\left[1+\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)+\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)^2+\dots\right]\\
                &=t^{-1}+\frac{1}{3!}t+\left[\left(\frac{1}{3!}\right)^2-\frac{1}{5!}\right]t^3+\dots
\end{align}
Fair enough. However, I'm not sure how the person motivates that $|\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots| < 1$ (which is necessary for the above calculation to be valid, i.e. the part of writing $\frac{1}{1-\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots}$ as a geometric series).
This may be obvious but I cant seem to figure it out. It feels valid to say that it should be less than one, but how do you show this?
Thanks!

Here is a simple justification:
$$1-\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}-\dotsm\right)=\frac{\sin t}t,$$
so that $\enspace\dfrac{t^2}{3!}-\dfrac{t^4}{5!}+\dfrac{t^6}{7!}-\dotsm=1-\dfrac{\sin t}t$, and it is well-known the cardinal sine function
$$\operatorname{sinc}(t)=\begin{cases}\dfrac{\sin t}t&\text{if }t\ne 0\\1&\text{if }t=0\end{cases}$$
has values in $[0,1]\:$ for all $|t|\le\frac\pi2$.
