Let $a_1$ and $b_1$ be any two positive numbers. Let $\alpha_{n+1} = \frac{2 \alpha_n \beta_n}{\alpha_n + \beta_n}$ and $\beta_{n+1} = \sqrt{\alpha_n \beta_n}$. Show that both sequences converge and have the same limit.
From my readings I know that these sequences are the harmonic and geometric means and that they both converge to the harmonic-geometric mean. However, I have no clue where to start in my proof. Any help would be greatly appreciated.
Since $a_1 > 0$ and $b_1 > 0$, show by induction that $\alpha_n > 0$ and $\beta_n > 0$ for all $n\in \Bbb N$. By the AM-GM inequality, $$\alpha_{n+1}\beta_{n+1} = \alpha_{n+1}\sqrt{\alpha_n\beta_n} \le \alpha_{n+1}\frac{\alpha_n + \beta_n}{2} = \alpha_n \beta_n$$ for all $n\in \Bbb N$. Hence, the sequence $(\alpha_n\beta_n)_{n\in \Bbb N}$ is a monotonic decreasing sequence of positive real numbers. Therefore, there is a real number $L$ such that $\lim_{n\to \infty} \alpha_n \beta_n = L$. Thus $\beta_n \to \sqrt{L}$. Since $\alpha_n = \beta_{n+1}^2/\beta_n$ for all $n$ and $\beta_n$ is convergent, so is $\alpha_n$. The limit of $\alpha_n$ will then be $(\sqrt{L}^2)/\sqrt{L} = L/\sqrt{L} = \sqrt{L}$. Thus $\lim_{n\to \infty} \alpha_n = \lim_{n\to \infty} \beta_n$.