Proving two sequences converge to the same limit $a_{n+1}\frac{a_n+b_n}{2} \ , \ b_{n+1}=\frac {2a_nb_n}{a_n+b_n} $ 
$\text{We have two sequences}$ $(a_n), (b_n)$ where $0<b_1<a_1$ and: 
$$a_{n+1}=\frac{a_n+b_n}{2} \ , \ b_{n+1}=\frac {2a_nb_n}{a_n+b_n} $$
Prove both sequences converge to the same limit and try to find the
  limit.

What I did: Suppose $\displaystyle\lim_{n\to\infty}a_n=a, \displaystyle\lim_{n\to\infty}b_n=b$ So $\displaystyle\lim_{n\to\infty} \frac {a_n+b_n} 2= \frac{a+b} 2 =K$
Take $a_{n+2}= \frac {a_{n+1}+b_{n+1}}{2}=\frac {\frac{a_n+b_n}{2}+\frac {2a_nb_n}{a_n+b_n}}{2}=...=X$
We know that as $n$ tends to infinity $\lim x_n= \lim x_{n+1}$ so: $X=K$ and after some algebra I get $a=b$
As for the limit, it depends on only one of the sequences, since both tend to the same limit. The limit can be any constant or $\pm\infty$. 
Is this approach correct ?
I excluded the algebra because I type this manually and to make the solution easier to read. 

Firstly, in your initial solution K = a by definition, so the result a=b follows immediately.
For an estimate of the limit for large n observe that the next member of each sequence is between an and bn, and so the final result must also be between a1 and b1, and indeed between (a1 + b1)/2 and 2*(a1*b1)/(a1+b1), and so certainly not infinity.
