let $t$ be a real number .
show that : $\operatorname{cosh}(t) \leq e^{\frac{t^2}{2}}$
it was imposed to use power series of the two functions to show it .
I have for each real $t$ :
$\operatorname{cosh}(t)=\sum_{n=0}^{\infty} \frac{ t^{2n}}{(2n)!}$
$e^{ \frac{t^2}{2}}= \sum_{n=0}^{\infty} \frac{ t^{2n}}{2^nn!}$
1- what am I supposed to remark ?
2- In general , is using power series a good method to show inequalities ?
Edit:
as a hint given to me :
I ll show that:
$(2n)! \leq 2^n n$
using recurrence i have $(2(n+1))!=(2n)!(2n+1)(2n+2) \leq 2^n n(2n+1)(2n+2) \leq 2^{n+1} (n+0,5)(n+1) \leq 2^{n+1}(n+1)$
We need to prove that $$\frac{e^t+e^{-t}}{2}\leq e^{\frac{t^2}{2}}$$ or $f(t)\geq0,$ where $$f(t)=\frac{t^2}{2}+t+\ln2-\ln(e^{2t}+1).$$ But $$f''(t)=\left(\frac{e^{2t}-1}{e^{2t}+1}\right)^2\geq0,$$ which says that $f'$ increases.
Also, $$f'(t)=t+1-\frac{2e^{2t}}{e^{2t}+1},$$ which gives $f'(0)=0$ and for $t=0$ $f$ gets a minimal value.
Id est, $$f(t)\geq f(0)=0$$ and we are done!
Also, you can use that $f(0)=f'(0)=0$ and $f''(t)\geq0$ for all real $t$ and $f''(\theta)\geq0.$