Prove $\sum\limits_{n=0}^\infty\frac{5^n(3^{5^{n+1}}-5\cdot3^{5^n}+4)}{(729)^{5^n}-(243)^{5^n}-5\cdot3^{5^n}+1}=\frac12$ This problem is taken from Algerian Olympiad and asks to prove that
$$\sum_{n=0}^{\infty} \dfrac{5^n(3^{5^{n+1}} -5\cdot3^{5^n} + 4)}{(729)^{5^n} - (243)^{5^n}-5\cdot3^{5^n}+1} = \frac 12.$$
Noticing that $729=3^6$ and $243 = 3^5$, I tried simplifying the general terms by setting $x=3^{5^n}$ but it seems to give no simplifications.
Thanks in advance for any advice / ideas.

The series does indeed converge (by the ratio test, for example), but its sum is not $\,\dfrac{1}{2}\,$, it is rather strictly greater than $\,\dfrac{1}{2}\,$. The first two terms add up to  $\dfrac{29}{59}$ $+$ $\dfrac{529555380145}{25630480435499}$ $\simeq 0.512$ $\gt \dfrac{1}{2}\,$ already, and adding more positive terms will only increase the sum further.
To show that all terms are in fact positive, it is enough to note that with $\,x=3^{5^n} \ge 3\,$ both:


*

*numerator $\,5^n(x^5-5x+4) = 5^n(x-1)^2(x^3+2x^2+3x+4) \ge 0$;

*denominator $\,x^6-x^5-5x+1 = u^6 + 11 u^5 + 50 u^4 + 120 u^3 + 160 u^2 + 107 u + 23\ge 0\,$ where $\,u=x-2 \ge 0\,$.
