Proof that $ \pi > 2 \cdot \sqrt{2}$ and $ \pi > 3 $ I need to prove that $\pi > 3$ and  $ \pi > 2 \cdot \sqrt{2}$ only in use of definition of cosine (by series) or $\cos(x) = \frac{e^{iz}+ e^{-iz}}{2}$and definition of $\pi$ as $\pi = 2\cdot x_0$ where $cos(x_0) = 0$ and $x_0 \in (0,2)$
what I did
I thought that I can use $\frac{\pi^2}{6} = \sum\frac{1}{n^2}$:
$$\frac{\pi^2}{6} = \sum\frac{1}{n^2} > 1 + 1/4 + 1/9 + 1/16 = \frac{205}{144}$$
$$ \pi^2 > \frac{205}{24} > 8 $$
$$ \pi > 2 \cdot \sqrt{2} $$
Fine... but there are 2 problems: firstly, in this way is hard to proof that $\pi > 3$. Moreover I have just understood that  I can't use $\sum\frac{1}{n^2}$ because "it was not main part of my lecture"

The area of a regular octagon is $2\sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $\pi\gt 2\sqrt{2}$.
Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $\pi\gt 3$.

Approach using the series definition of cosine and pi.
We can use the fact that $\cos(x)$ is continuous, and that
$$\cos(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$$
We have that $\cos(x)\gt 0$ for $x\lt \pi/2$ and $\cos(x)\lt 0$ for $x\gt \pi/2$ (assuming that $x\in (0,2)$). Consider the sum
$$\sum_{n=0}^\infty \frac{(-1)^n(9/4)^{n}}{(2n)!}$$
By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
$$\frac{3245071}{45875200}\approx 0.0707$$
with an error less than $10^{-6}$. This shows that $\cos(3/2)\gt 0$ and $\pi\gt 3\gt 2\sqrt{2}$.
