<s>[INST] Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$. Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$.
This inequality came up
in my solution to
Show that the sequence $\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$ converges.
This is in the category of
"There should be
a number of ways
to prove this."
Here is one way
I came up with.
I am interested in seeing
how many others
there are.
$(\ln(\cos(x))'
=-\tan(x)
\le -x$
for $0 \le x \lt \pi/2$.
Integrating from $0$ to $x$,
since the two sides are $0$
at $x=0$,
$\ln(\cos(x))
\le -x^2/2$
so
$\cos(x)
\le e^{-x^2/2}
$.
 [/INST] By Taylor's expansion
$$\cos(x) \le 1-\frac{x^2}2+\frac{x^4}{24} \le 1-\frac{x^2}2+\frac{x^4}{8}-\frac{x^6}{48} \le e^{-x^2/2}$$
indeed
$$1-\frac{x^2}2+\frac{x^4}{24} \le 1-\frac{x^2}2+\frac{x^4}{8}-\frac{x^6}{48}\iff \frac{x^4}{12}-\frac{x^6}{48}\ge 0 \iff x^4(4-x^2)\ge 0$$ 
</s>