<s>[INST] Series Convergence of Harmonic Means 
Let $\{x_n\}$ be a sequence of real numbers such that $0< x_1 <x_2$. If $$x_n=  \frac{2}{\frac{1}{x_{n-1}}+\frac{1}{x_{n-2}}}, $$then show that 
  $$\lim_{n\to\infty}x_n=\frac{3x_1x_2}{2x_1+x_2}.$$

 [/INST] Note that
\begin{align}
& x_n = \frac{2}{\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}} \\
\iff & \frac{1}{x_{n-1}} + \frac{1}{x_{n-2}} = \frac{2}{x_{n}} \\
\iff & a_n = \frac{1}{2}a_{n-1} + \frac{1}{2}a_{n-2}\\
\iff & a_n-a_{n-1} = -\frac{1}{2}\left(a_{n-1}-a_{n-2}\right)
\end{align}
where $a_i = \frac{1}{x_i}$. Consequently, 
$$\sum_{n=3}^{k}(a_n-a_{n-1}) = a_k-a_2 = -\frac{1}{2}\sum_{n=3}^{k}(a_{n-1}-a_{n-2}) = -\frac{1}{2}(a_{k-1}-a_1)$$
Let $\alpha = \lim_{k\to \infty} a_k$, then the above equation implies
$$\alpha -a_2 =-\frac{1}{2}(\alpha-a_1) \implies \alpha = \frac{2a_2+a_1}{3}.$$
Therefore, $$\lim_{n\to\infty}x_n = \frac{1}{\alpha} = \frac{3x_1 x_2}{2x_1 + x_2}.$$
</s>