I'm reading this survey by Carl Offner about digit computation of the number $\pi$.
In page 7 there's a step that I didn't understand:
suppose
$$\alpha_{n+1}=\frac{\alpha_n \beta_n}{\alpha_n + \beta_n}$$ $$\beta_{n+1}=\sqrt{\frac{ \beta_n \alpha_{n+1}}{2}}$$
and
$$a_n=6\cdot 2^n \alpha_n$$ $$b_n=6\cdot 2^n \beta_n$$
Why it "follows easily" then, that
$$a_{n+1}=\frac{2a_nb_n}{a_n+b_n}$$ $$b_{n+1}=\sqrt{b_na_{n+1}}$$
I've tried computing
$$a_{n+1}=6\cdot 2^{n+1}\alpha_{n+1}$$
and then plugging the expression for $\alpha_{n+1}$ but it doesn't really help.
Thanks for your help!
Write $\alpha_n = \frac{a_n}{6\cdot 2^n}$ and $\beta_n = \frac{b_n}{6\cdot 2^n}.$ Then plug them into $$\alpha_{n+1}=\frac{\alpha_n \beta_n}{\alpha_n + \beta_n}$$
Get $$\frac{a_{n+1}}{6\cdot 2^{n+1}}=\frac{\frac{a_n}{6\cdot 2^n} \frac{b_n}{6\cdot 2^n}}{\frac{a_n}{6\cdot 2^n} + \frac{b_n}{6\cdot 2^n}} = \frac{a_n b_n }{a_n+b_n} / 6\cdot 2^n$$ So $$a_{n+1}=\frac{2a_nb_n}{a_n+b_n}.$$
Same thing for the second equation.