Tweaking an easy problem

Define $g(n):=\phi_1^n-\phi_2^n$ . These functions have a nice Fibonacci-like property:

$g(n-1)+g(n) =\phi_1^{n-1}-\phi_2^{n-1}+\phi_1^n-\phi_2^n\\ =\phi_1^{n-1}(\phi_1+1)-\phi_2^{n-1}(\phi_2+1)$

$=\phi_1^{n-1}\phi_1^2-\phi_2^{n-1}\phi_2^2\\ =\phi_1^{n+1}-\phi_2^{n+1}\\ =g(n+1).$

This allows us to write:

$\sum_{n=2}^\infty \frac{\phi_1^n-\phi_2^n}{\phi_1^{2n}+\phi_2^{2n}-\phi_2^{n-1}\phi_1^{n-1}(\phi_2^2+\phi_1^2)}\\ =\sum_{n=2}^\infty \frac{\phi_1^n-\phi_2^n}{(\phi_1^{n+1}-\phi_2^{n+1}) (\phi_1^{n-1}-\phi_2^{n-1})}$ $=\sum_{n=2}^\infty \frac{g(n)}{g(n-1)g(n+1)}\\ =\sum_{n=2}^\infty \frac{g(n+1)-g(n-1)}{g(n-1)g(n+1)}\\ =\sum_{n=2}^\infty \frac1{g(n-1)}-\frac1{g(n+1)}$ which telescopes into $\frac1{g(1)}+\frac1{g(2)}=\frac1{\sqrt5}+\frac1{\sqrt5}=\frac2{\sqrt5}$ , giving us $a=2$ and $b=5$ .

The second part of the exercise is to analyse $f$ ; in this case, we have:

$f(2,5)=\prod_{i=0}^\infty \left(1+\frac1{5^{2^i}}\right).$

Observe that:

$\left(1-\frac15\right)f(2,5) =\left(1-\frac15\right)\left(1+\frac15\right) \left(1+\frac1{5^2}\right) \left(1+\frac1{5^4}\right)\left(1+\frac1{5^8}\right)\dots$

$=\left(1-\frac1{5^2}\right) \left(1+\frac1{5^2}\right) \left(1+\frac1{5^4}\right)\left(1+\frac1{5^8}\right)\dots$

$=\left(1-\frac1{5^4}\right)\left(1+\frac1{5^4}\right)\left(1+\frac1{5^8}\right)\dots$

$\vdots$

$=1$ ,

seeing as we get $1-\frac1{5^{2^n}}$ with steadily increasing $n$ , which converges to $1$ . Hence $f(2,5)=\frac1{1-\frac15}=\frac54$ , therefore the answer is $\alpha+\beta=5+4=\boxed{9}$ .