Estimating pi Archimedes method I'm self-teaching real analysis. I found the following (lengthy) question on some old lecture notes, and I'm a bit stuck. I've posted what I've attempted below as well as the question.
Let $A_{n}$ and $a_{n}$ be the area of the circumscribed and inscribed regular n sided polygon of radius 1, respectively.
Archimedes' used the formula:


*

*$A_{2n} = \frac{2A_{n}a_{2n}}{A_{n}+a_{2n}}$

*$a_{2n} = \sqrt{a_{n}A_{n}}$
Using trig, I arrive at the following:


*

*$A_{n} = n\tan(\frac{\pi}{n})$

*$a_{n} = \frac{n}{2}\sin(\frac{2\pi}{n})$
Question: Why is the sequence $a_{4}, a_{8},  a_{16}, a_{32} \dots$ increasing? Furthermore why are all the values between $2$ and $\pi$? What similar statement can be made about $A_{4}, A_{8}, A_{16}, A_{32} \dots$?
Question: Why can $\Big(\frac{a_{2n}\sqrt{A_{n}}}{(A_{n}+a_{2n})(\sqrt{A_{n}}+\sqrt{a_{n}})}\Big)$ never be larger than $0.4$? Show that the error ($A_{n}-a_{n})$ in calculating $\pi$ reduces by at least $0.4$ when $n$ is replaced by $2n$. 
Question: Finally, show that by calculating $A_{2^{10}}$ and $a_{2^{10}}$ we can estimate $\pi$ to within $0.0014$
Attempt at Question 1: $a_{4}= 2, a_{8} = 2\sqrt{2}, \dots$. The sequence is an increasing sequence, because, in each subsequent term, the previous $n$ is used in it. (e.g. $a_{8} = a_{4}\sqrt{2}$). Therefore, for all $n$ we have $a_{n} \leq a_{2n}$ which by definition means the sequence is increasing. 
The values are all between $2$ and $\pi$ because, we know that $a_{n} \leq \pi \leq A_{n}$ and the lower bound of the sequence is $2$. 
The similar statement about $A_{4}, A_{8} \dots$ is that $A_{4} = 4, A_{8} = 8(\sqrt{2}- 1), \dots$. The sequence is a decreasing sequence. [I'm not sure how to explain why qualitatively]. 
The values are all between $4$ and $\pi$ because we know that $a_{n} \leq \pi \leq A_{n}$ and the upper bound of the decreasing sequence is $4$.
Attempt at Question 2:I'm stuck. I have found that the given expression is derived by taking $A_{2n} - a_{2n}$. So I assume I have to use the bounds I derived in Question 1 (perhaps doubling them since the bound in question 1 were for $A_{n}$ and $a_{n}$)?
Attempt at Question 3: I took $n=2^{10}$ and substituted this into the formula for $a_{n}$ and $A_{n}$ respectively. This gives
$a_{2^{10}} = 3.14157294, A_{2^{10}} = 3.14160251$. The error is given by $(A_{n}-a_{n})$ therefore taking the difference I get an error of
$2.95702568 \times 10^{-5}$, which is within the required amount.
 A: For the lower sequence:
$$ a_{4n} = \sqrt{a_{2n}A_{2n}} = a_{2n}\sqrt{\frac{2A_n}{A_n + a_{2n}}} $$
Saying that "it uses the value before it" is not really precise in the way you stated it, in particular the example is rather misleading. While it is true that the $a_{2n}$ appears outside the root, it also appears inside the root. One has to make sure that this doesn't affect the overall behaviour of the sequence. Luckily this is the case, since:
$$ a_{2n} < A_n \implies \sqrt{\frac{2A_n}{a_{2n} + A_n}} > \sqrt{\frac{2A_n}{A_n + A_n}} = 1 \implies a_{4n} > a_{2n} $$
Likewise for the upper sequence we have:
$$ A_{2n} = \frac{2 A_n a_{2n}}{A_n + a_{2n}} < \frac{2 A_n a_{2n}}{a_{2n} + a_{2n}} = A_n $$
Now the bound for the expression is actually even lower:
$$ \frac{a_{2n}\sqrt{A_n}}{(a_{2n} + A_n)(\sqrt{A_n} + \sqrt{a_n})} = \frac{a_{2n} A_n}{(a_{2n} + A_n)^2} \leq \frac 1 4 $$
This is a rather well-known inequality that holds for all positive real numbers.
A: @Niki Di Giano thanks for the detailed answer. I'm not quite following how you get $\frac{a_{2n}\sqrt{A_{n}}}{(a_{2n}+A_{n})(\sqrt{A_{n}}+\sqrt{a_{n}})} = \frac{a_{2n}A_{n}}{(a_{2n}+A_{n})^{2}} \leq \frac{1}{4}$.   I rearranged $a_{2n} = \sqrt{a_{n}A_{n}}$ to get $\frac{a^{2}_{2n}}{A_{n}} = a_{n}$ and substituted this into the denominator. Then rationalised the denominator. Is this going in the right direction?
