Can someone give an analytical proof that $\pi$ is greater than 3?

Question

The geometrical proof is simple, but I want to know how to prove $\pi \gt 3$ by using basic calculus?

Answer 1

There is a feature column in AMS titled Calculating Pi using Elementary Calculus that you might want to check out.

Answer 2

$$\sin(x)<x$$ whenever $x>0$. Take $x=\pi/6$.

ADDED IN EDIT

Apologies to Spine Feast who just pipped me to this, but $$\frac\pi6=\int_0^{1/2}\frac{dt}{\sqrt{1-t^2}}>\int_0^{1/2}dt=\frac12$$ is the same thing again but even more analytical.

Answer 3

I'm sure there are many, but here is a proof that uses the famous infinite sum

$$ \sum_{n=1}^\infty\frac{1}{n^2} = \frac{\pi^2}{6} $$ Solving for $\pi$ and using the fact that an infinite sum of positive terms increases monotonically to its limit, we see that

$$ \pi = \left(6\sum_{n=1}^\infty\frac{1}{n^2}\right)^{1/2} > \left(6\sum_{n=1}^7\frac{1}{n^2}\right)^{1/2}> \sqrt{9} = 3 $$ You can see that $6\sum_{n=1}^7 \frac{1}{n^2} > 9$ by direct calculation.

Answer 4

The integral

$$\int_0^1 \frac{x(1-x)^2}{1+x^2}dx$$

evaluates to $\frac{\pi-3}{2}$ and the result is positive because the integrand is $\ge0$ in the interval $(0,1)$.

A related series is $$\pi = 3+ \sum_{k=0}^\infty \frac{24}{(4k+2)(4k+3)(4k+5)(4k+6)}$$