convergence of sequence defined recursively $\frac{2}{a_{n+2}}=\frac{1}{a_{n+1}}+\frac{1}{a_n}$ How to prove that the sequence ${a_n}$ defined by $\frac{2}{a_{n+2}}=\frac{1}{a_{n+1}}+\frac{1}{a_n}$ for $n\geq1$ and $0\lt{a_1}\lt{a_2}$ converges? How to find its limit?  I truly do not know how to proceed further. I tried to use A.M>G.M inequality but with no success. 

As it was suggested in the comments use $b_n=\frac{1}{a_n}$ and then solve $$2b_{n+2}=b_{n+1}+b_n$$
using characteristic polynomials technique (or look at this similar question), leading to the following polynomial
$$2x^2-x-1=0$$
with solutions $x_1=1$, $x_2=-\frac{1}{2}$ and general solution for the recurrence
$$b_n=Ax_1^n+Bx_2^n=A+B\left(-\frac{1}{2}\right)^n \Rightarrow 
a_n=\frac{1}{A+B\left(-\frac{1}{2}\right)^n}$$
and $\lim\limits_{n\rightarrow\infty} a_n = \frac{1}{A}$. Given $a_1=\frac{1}{A-\frac{B}{2}}$ and $a_2=\frac{1}{A+\frac{B}{4}}$, it's not difficult to find $A$:
$$A-\frac{B}{2}=\frac{1}{a_1} , 2A+\frac{B}{2}=\frac{2}{a_2} \Rightarrow 
A=\frac{1}{3a_1}+\frac{2}{3a_2}$$
