<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] Solution
Since $$x_n = \frac{2}{\dfrac{1}{x_{n-1}} + \dfrac{1}{x_{n-2}}},$$
then $$ \frac{2}{x_{n}}=\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}.$$
Thus, $$2y_n=y_{n-1}+y_{n-2},$$ where $y_n=\dfrac{1}{x_n}$.
From $(1)$, we may obtain $$y_n-y_{n-1} = -\frac{1}{2}\left(y_{n-1}-y_{n-2}\right),$$ for $n=3,4,\cdots.$
Hence，$$y_n-y_{n-1}=(y_2-y_1)\left(-\frac{1}{2}\right)^{n-2}.$$
Thus, $$y_n-y_1=\sum_{k=2}^n(y_k-y_{k-1})=\sum_{k=2}^n(y_2-y_1)\left(-\frac{1}{2}\right)^{k-2}=(y_2-y_1)\cdot \frac{2}{3}\left[1-\left(-\frac{1}{2}\right)^{n-1}\right]$$
Thus, $$y_n=y_1+(y_2-y_1)\cdot \frac{2}{3}\left[1-\left(-\frac{1}{2}\right)^{n-1}\right].$$
Let $n \to \infty$. Then$$y_n \to y_1+(y_2-y_1)\cdot \frac{2}{3}=\frac{1}{3}y_1+\frac{2}{3}y_2=\frac{1}{3x_1}+\frac{2}{3x_2}=\frac{2x_1+x_2}{3x_1x_2}.$$
As a result, $$x_n=\frac{1}{y_n} \to \frac{3x_1x_2}{2x_1+x_2}.$$
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