<s>[INST] Algebraic proof of $\tan x>x$ I'm looking for a non-calculus proof of the statement that $\tan x>x$ on $(0,\pi/2)$, meaning "not using derivatives or integrals." (The calculus proof: if $f(x)=\tan x-x$ then $f'(x)=\sec^2 x-1>0$ so $f$ is increasing, and $f(0)=0$.) $\tan x$ is defined to be $\frac{\sin x}{\cos x}$ where these are defined by their infinite series. What I have so far:
$$|z|\le1\implies\left|\sum_{n=4}^\infty\frac{z^n}{n!}\right|<\sum_{n=0}^\infty\frac{|z|^4}{4!\,5^n}=\frac{5|z|^4}{4\cdot 4!}$$
$$\left|\sin x-\Big(x-\frac{x^3}6\Big)\right|=\Im\left[\sum_{n=4}^\infty\frac{(ix)^n}{n!}\right]<\frac{5x^4}{4\cdot 4!}<\frac{x^3}6$$
$$\left|\cos x-\Big(1-\frac{x^2}2\Big)\right|=\Re\left[\sum_{n=4}^\infty\frac{(ix)^n}{n!}\right]<\frac{5x^4}{4\cdot 4!}<\frac{x^2}6$$
Thus $\sin x>x-\frac{x^3}3$ and $\cos x<1-\frac{x^2}3$, so $\tan x>x$. However, this only covers the region $x\le1$, and I still need to bound $\tan x$ on $(1,\pi/2)$. My best approximation to $\pi$ is the very crude $2<\pi<4$, derived by combining the above bounds with the double angle formulas (note that $\pi$ is defined as the smallest positive root of $\sin x$), so I can't quite finish the proof with a bound like $\sin x>1/\sqrt 2$, $\cos x\le\pi/2-x$ (assuming now $x\ge1\ge\pi/4$) because the bound is too tight. Any ideas?
 [/INST] (I should have added this as comment but don't have privilege yet)
Just in case OP haven't checked out yet, the infinite series expansion of $tan(x)$ is already available on Wikipedia. Quoting here:
$$tan(x) = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 + {...}$$
for $\lvert x \rvert < \frac{\pi}{2}$.
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