Proving convergence of a recursively defined sequence Using a computer it is easy to see that the sequences defined by letting $a_1=1$, $a_2=m$, and
$$a_n=\frac{2a_{n-1}a_{n-2}}{a_{n-1}+a_{n-2}}$$
converges to $\frac{3m}{m+2}$. I would very much like to know how to prove this. I don't even know where to start.
Hint: The following is I think a natural starting "move."  

Let $a_n=\dfrac{1}{x_n}$.  When the smoke clears you get a nice linear recurrence. 

Continuation: Making the suggested substitution, and simplifying, we obtain the recurrence $x_n=\dfrac{1}{2}\left(x_{n-1}+x_{n-2}\right)$. We have to be a little careful, since there will be trouble if $a_{n-1}+a_{n-2}=0$. But for example if $m\gt 0$ then all terms are safely positive. 

There are many ways to solve the recurrence. For example we can use a general technique for solving homogeneous linear recurrences with constant coefficients, and write down the characteristic polynomial $2x^2-x-1=0$, which has the roots $1$ and $-\dfrac{1}{2}$. 

Thus the general solution of the recurrence is $x_n=A+B\left(-\frac{1}{2}\right)^n$. Putting $x_1=1$ and $x_2=\dfrac{1}{m}$ we find that $A=\dfrac{m+2}{3m}$, and  $B=\dfrac{4-2m}{3m}$. Thus
$$x_n=\frac{m+2}{3m} + \left(-\frac{1}{2}\right)^n \frac{4-2m}{3m}.$$
As $n\to\infty$, $x_n\to \dfrac{m+2}{3m}$. 

Remark: Because the numbers are quite simple, we do not need to use the "general" procedure to solve the recurrence. More generally, consider the recurrence
$$y_{n}=py_{n-1}+(1-p)y_{n-2},$$
which comes up in some probability calculations.
This may be rewritten as
$$y_n-y_{n-1}=(-1)(1-p)(y_{n-1}-y_{n-2}).$$
If we let $z_n=y_{n}-y_{n-1}$, we get $z_n=(-1)(1-p)z_{n-1}$. So the sequence $(z_n)$ is a geometric sequence with common ratio $-(1-p)$. Now we can easily find a general expression for $z_n$, and hence for $y_n$.
Let $A_n=\frac{a_n}{a_{n-1}}$, then we have $A_n=\frac{2}{1+A_{n-1}}$, it is easy to find that $\lim\limits_{n\to\infty }A_n$ exists 
then we have $\lim\limits_{n\to\infty }\frac{a_n}{a_{n-1}}=\frac{\sqrt{2}-1}{2}&lt;1$, we can use the $\epsilon- N$ to illustrate the result......