Showing $\cos \left( \frac{1}{n} \right) > 1 - \frac{1}{2n^2}$ I'm interested in showing the inequality
$$ \cos\left(\frac{1}{n}\right) > 1-\frac{1}{2n^2}$$
For all $n$ positive integer. I've tried to proceed by induction and it's easily seen (assuming properties of trigonometric functions) that it holds for $n=1$ but I get stuck on showing it for $n+1$. Once $\cos(x)$ is decreasing on $[0,1]$ we have $$\cos\left(\frac{1}{n+1}\right) > \cos\left(\frac{1}{n}\right) > 1- \frac{1}{2 \, n^2}.$$ But I don't know how to get it from here.
 A: $$\cos x\le1$$ and equality holds at $x=0$. Integrating from $0$ to $x$,
$$\sin x\le x$$
and equality holds at $x=0$. Integrating from $0$ to $x$,
$$-\cos x\le \frac{x^2}2-1.$$
Take $x=\dfrac1n$.

You can continue forever and retrieve the Taylor series.

A: Put
$$f(x)=\cos(x)+\frac{x^2}{2}-1$$
$$f(0)=0$$
for $x\in(0,1]$, $f$ is continuous at $[0,x]$ and differentiable at $(0,x)$ thus by MVT
$$f(x)=xf'(c_1)=x(c_1-\sin(c_1))$$
with $$0<c_1<x\le 1$$
but
$$\sin(c_1)=\sin(c_1)-\sin(0)=c_1\cos(c_2)$$
with  $0<c_2<c_1<x\le 1$ and $\cos(c_2)<1$.
thus $$\sin(c_1)<c_1$$
and finally
$$f(x)>0$$
now replace $x$ by $\frac 1n$.
A: Let consider $f(x)=\cos x-1+x^2$ which is even then we can consider $x\ge 0$ and we have
$$f(x)=\cos x-1+x^2\ge 0 \quad f(x)=0 \iff x=0$$
indeed
$$g(x)=f'(x)=x-\sin x\ge 0$$
indeed
$$g'(x)=f''(x)=1-\cos x\ge 0 \quad g'(x)=0 \iff x=k\frac{\pi}2$$
therefore for $x\neq 0$
$$\cos x-1+x^2>0 \implies \cos\left(\frac{1}{n}\right) > 1-\frac{1}{2n^2}$$
A: Although there is already an accepted answer I'd like to add a geometric proof of the inequality:

$$x^2 > s^2 = (1-\cos^2 x) + (1-\cos x)^2 = 2(1- \cos x) \Rightarrow \boxed{\cos x > 1- \frac{x^2}{2}}$$
