Proof that $ \pi > 2 \cdot \sqrt{2}$ and $ \pi > 3 $

Question

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"

Sum up to the temr $1/7^2$ and you'll get $\pi > 3$. And if you prove $\pi > 3$ then $\pi > 2\sqrt{2}$ follows since $3 > 2\sqrt{2}$. — Winther, Dec 18 '18 at 22:48
Prove that $\pi>3$ using geometry ; Can someone give an analytical proof that $\pi$ is greater than 3? — Winther, Dec 18 '18 at 22:52
@Winther I saw this post. I use browser before I ask. There are solution in use of other ways. Not what I am looking for — , Dec 18 '18 at 22:55
Why don't you use the power series and plug in $x = \sqrt{2}$ or $x = 3/2$ and compute the partial sums? — minimalrho, Dec 18 '18 at 22:59

Franklin Pezzuti Dyer · Accepted Answer · 2018-12-19T21:16:48.703

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 $\pi\gt 3\gt 2\sqrt{2}$.

@VirtualUser I have edited to show you how to do it using the series definition of $\pi$. — Franklin Pezzuti Dyer, Dec 18 '18 at 23:04

score 1 · Answer 2 · answered Dec 18 '18 at 23:10

You can show that in $[0,2]$n the cosine function is decreasing. Then

$$\frac\pi2>\frac32\iff0<\cos\frac32.$$

You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).

The first partial sums are $$1,-\frac18,\frac{11}{128},\frac{359}{5120},\frac{16229}{229376},\frac{3245071}{45875200},\cdots$$

Below, the last partial sum above, and the exact value. $$0.0707369341169\leftrightarrow 0.0707372016677$$

Proof that $ \pi > 2 \cdot \sqrt{2}$ and $ \pi > 3 $

what I did

2 Answers2