estimation of $\pi$ and $e$ by using the Taylor series of $\cos x$ how can one show, that $3<\pi<3.2$, 
$2.7<e<3$ by just knowing, the estimation of $\cos(x)$, namely:
$$1-x^2/2+x^4/24-x^6/720\le\cos(x)$$  ?
If I substitute Pi/2 into this estimation, I just have an expression with a lot of Pies..?
Do I have to use the Euler's formula for the second exercise?
Thank you
 A: I'll show how to use this to get $3.1 < \pi$ with no assumptions and $\pi < 3.2$ if we assume we already know it's less than 6.

Showing $\pi < 3.2$:
$1 - (\pi/2)^2/2 + (\pi/2)^4/24 \geq \cos(\pi/2) = 0$
Write $x = (\pi/2)^2$
Then $x^2/24 - x/2 + 1 \geq 0$.
This is quadratic, minimum at 6, and negative at both 9 and at 2.54 (hence for all values between). Assume $\pi < 6$ so $\pi/2 < 3$ and $(\pi/2)^2 < 9$. Then $(\pi/2)^2 < 2.54$. Now $(3.2/2)^2 = 2.56$ so $\pi < 3.2$.

Showing $\pi > 3.1$:
Now consider $1 - (\pi/2)^2/2 + (\pi/2)^4/24 - (\pi/2)^6/720 \leq 0$.
Again write $x = \pi/2$. Then:
$1 - x/2 + x^2/24 - x^3/720 \leq 0$
Claim: This cubic is strictly decreasing.
Proof: Take the derivative. It's $-1/2 + x/12 - x^2/120$. This has a maximum at $x = 5$, where it's negative. Thus the cubic is strictly decreasing.
Now we can just check some values of the cubic. We see when $x = 2.46$ it's positive. Thus $(\pi/2)^2 > 2.46$.
Now $(3.1/2)^2 = 2.4025 < 2.46$ so $\pi > 3.1$.

Supplement showing $\pi < 6$:
There's no chance we can show $\pi < 6$ like this, because all we're using is that $\cos(\pi/2) = 0$, but cosine is also zero at $3\pi/2$ etc.
To see $\pi < 6$, there are various methods. For example, use the geometric proof that $\pi < 4$ because the unit circle is inscribed in a square of side length two, so the area of the unit circle, $\pi$, is less than the area of the square, 4.
