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. 
 A: Note that $2\left(\frac{1}{a_{n+2}}-\frac{1}{a_{n+1}} \right)=-\left(\frac{1}{a_{n+1}}-\frac{1}{a_{n}}\right)$, thus with $b_n = \frac{1}{a_{n+1}}-\frac{1}{a_{n}}$, $b_{n+1}=-\frac 12 b_n$.
Hence $b_{n+1} = \frac{(-1)^n}{2^n}b_1$, so $\sum_n b_n$ converges, thus $\frac1{a_n}$ converges, and so does $a_n$.
Let $\ell=\lim_n a_n$. Then  $\sum_{n=1}^\infty b_n = \frac 1\ell -\frac 1{a_1}$, which rewrites as $\frac 23 \left(\frac 1{a_1} - \frac 1{a_2} \right) = \frac 1\ell -\frac 1{a_1}$, hence $$\ell = \frac{1}{\frac 23 \left(\frac 1{a_1} - \frac 1{a_2} \right)+\frac 1{a_1}}$$
A: 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 $A$:
$$A-\frac{B}{2}=\frac{1}{a_1} , 2A+\frac{B}{2}=\frac{2}{a_2} \Rightarrow 
A=\frac{1}{3a_1}+\frac{2}{3a_2}$$
A: Too involved for a comment: Another way to prove the limit exists:
For positive $a, b$ let $H(a,b)=(\frac {1}{2}(a^{-1}+b^{-1}))^{-1}.$ We have $a_{n+2}=H(a_{n+1},a_n).$
We have $0< a\leq b\implies a\leq H(a,b)\leq b.$ Therefore   the sequence $S=(a_{2n-1})_{n\in \Bbb N}$ is increasing and the sequence $T=(a_{2n})_{n\in \Bbb N}$ is decreasing, and every term of $S$ is $\leq$ every term of $T$. So $S$ converges to a limit $D$ while $T$ converges to a limit $U,$ with $0<a_1\leq D\leq U\leq a_2.$ 
Observe that $H:\Bbb R^+\times \Bbb R^+\to \Bbb R^+$ is continuous, so $$D=\lim_{n\to \infty}a_{2n+1}=$$ $$=\lim_{n\to \infty}H(a_{2n}, a_{2n-1})=$$ $$=H(\lim_{n\to\infty}a_{2n},\lim_{n\to \infty}a_{2n-1})=H(U,D).$$  But for any positive $U,D$ we have $H(U,D)=D \iff U=D.$
A: 2/a[n+2]  = 1/a[n+1]  + 1/a[n]  , get common denominator
2/a[n+2] = [a(n+1) + a(n)]/[ a(n+1) . a(N) ] , invert 
a[n+2] = 2[a(n+1).a(n)]/[ a(n+1) + a(n) ]   eqn 1
a(2) = 2 a(1).a(0)/ [1a(1) +1a(0)]   eqn 2  ,  note 1+1=2^1
a(3) = 2 a(2).a(1)/[ a(2) + a(1)] eqn 3 now substitute eqn 2 into eqn 3 and get 
a(3) = 4a(1).a(0)/[1a(1) + 3a(0)]  eqn 4 , note 1+3=4=2^2
using similar substitutions 
a(4) = 8a(1).a(0)/[ 3a(1) +5a(0)] eqn 5 , and 3+5=8=2^3
a(5) = 16a(1).a(0)/[5a(1) +11a(0)]  eqn 6, and 5+11=16=2^4
The sums of the co efficients of the denominator = the co efficient of the numerator and the  co efficient of the previous a(0) becomes the co efficient of the next a(1) , so we will expect the next term
a(6) = 2^5.a(1).a(0)/[ 11a(1) +p.a(0)] where p= 2^5-11=21 , so
a(6) = 32 a(1).a(0) /[ 11a(1) + 21a(0)]
a(n) = 2^(n-1).a(1).a(0)/[ (2^(n-1)-p)a(1) +p a(0)]
At this stage divide all the terms of the numerator and the denominator by 2^(n-1) , and let n tend to infinity .
Then the limit of a(n) and 1/a(n) has a limit as n tends to infinity. 
