Can someone give an analytical proof that $\pi$ is greater than 3? The geometrical proof is simple, but I want to know how to prove $\pi \gt 3$ by using basic calculus?
 A: $$\sin(x)<x$$
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.
A: 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 
$$\pi = 3+ \sum_{k=0}^\infty \frac{24}{(4k+2)(4k+3)(4k+5)(4k+6)}$$
A: There is a feature column in AMS titled Calculating Pi using Elementary Calculus that you might want to check out.
A: 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. 
