Wait...this converges?

Say $x_n=f_n$ , $y_n=f_{n-1}$ for $n \ge1$ . Also define $F(x,y)=\left(\frac{1}{x}+\frac{1}{y},x \right)$ so that $\left(x_{n+1},y_{n+1}\right)=F\left(x_{n},y_{n}\right)$

We can now use some stability theory . The fixed points of the map $F$ are at $x=y=\pm \sqrt2$ .

The Jacobian of $F$ is $J(x,y)=\begin{pmatrix} \frac{-1}{x^2} &\frac{-1}{y^2} \\ 1 & 0 \end{pmatrix}$

Evaluating this at either fixed point we get $J(\pm \sqrt2,\pm \sqrt2)=\begin{pmatrix} \frac{-1}{2} &\frac{-1}{2} \\ 1 & 0 \end{pmatrix}$

whose determinant is $\frac12$ . Since this is less than one in absolute value, the fixed points are both stable.

Now since both $f_0$ and $f_1$ are positive, all $f_n$ are; so the sequence can only converge to $\boxed{\sqrt2}$ .

Here's a plot of the above map:

If indeed this function converges, at some point it must recursively define itself - i.e. $f_{n} = \frac{f_{n} + f_{n}}{f_{n} * f_{n}}$ (but of course at infinity). Letting $f_{n} = x$ , we solve the equation $x = \frac{x + x}{x * x}$ , or $x = \frac{2x}{x^{2}}$ . Multiplying both ides by $x^{2}$ we arrive at $x^{3} = 2x$ , which transposes and factors as $(x)(x^{2} - 2) = 0$ . So our possible solutions are $x = 0, ± \sqrt{2}$ . However, by definition, all our terms are positive since the product of two positive terms can never be negative or zero. Thus, we discard the $x = 0$ and $x = - \sqrt{2}$ solutions. So our final answer is $x = \sqrt{2}$ , and in the required form it is $x = 1 * \sqrt{2} + 0$ . So $a + b + c = 1 + 2 + 0 = \boxed{3}$ .