I'm getting a bit puzzled with regards to this question:

I'm not really sure where to start with this except to state that 0.4 is an upper bound of the sequence. That isn't really enough for the start bit and I can't see what the next step would be.Explain why $\displaystyle \frac{a_{2n}\sqrt{A_n}}{(A_{n}+a_2n)(\sqrt{A_n}+\s qrt{a_n})}$ is never larger than 0.4 given that:

$\displaystyle a_{2n}=\sqrt{a_nA_n}$

$\displaystyle A_{2n}=\frac{2A_na_{2n}}{A_n+a_{2n}}$.

Hence show that the error $\displaystyle (A_n-a_n)$ in calculating $\displaystyle \pi$ reduces by at least 0.4 when replacing n by 2n. Show that by calculating $\displaystyle A_{2^{10}}$ and $\displaystyle a_{2^{10}}$ we can estimate $\displaystyle \pi$ to within 0.0014.

Okay, this may help a lot but i've just worked out that $\displaystyle 2 \leq a_n \leq \pi \leq A_n$ where $\displaystyle A_n $ tends to $\displaystyle \infty $ from above. Therefore the first bit should go a little something like this:

$\displaystyle \frac{\pi \sqrt{\pi}}{(\pi+\pi)(\sqrt {\pi}+\sqrt {\pi})}=\frac{\pi \sqrt{\pi}}{4\pi \sqrt {\pi}}=\frac{1}{4} $

The big problem here is that it doesn't give the required 0.4.

I'm also not sure if thiis helps but $\displaystyle a_4=2$ and $\displaystyle A_4=4$.