Prove $\sum\limits_{n=0}^\infty\frac{5^n(3^{5^{n+1}}-5\cdot3^{5^n}+4)}{(729)^{5^n}-(243)^{5^n}-5\cdot3^{5^n}+1}=\frac12$

I think the "$5 \cdot 3^{5^n}$" in the denominator is a typo and it can be proved that$$ \sum_{n = 0}^\infty \frac{5^n (3^{5^{n + 1}} - 5 \cdot 3^{5^n} + 4)}{3^{6 \cdot 5^n} - 3^{5^{n + 1}} - 3^{5^n} + 1} = \frac{1}{2}. $$

In fact,$$ \sum_{n = 0}^\infty \frac{5^n (3^{5^{n + 1}} - 5 \cdot 3^{5^n} + 4)}{3^{6 \cdot 5^n} - 3^{5^{n + 1}} - 3^{5^n} + 1} = \sum_{n = 0}^\infty \left( \frac{5^n}{3^{5^n} - 1} - \frac{5^{n + 1}}{3^{5^{n + 1}} - 1} \right) = \frac{5^0}{3^{5^0} - 1} = \frac{1}{2}. $$

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 $\,u=x-2 \ge 0\,$.

The series is convergent but it does not converge to $0.5$.

This series has a high degree of convergence and it converges to $0.512186580754$ which is greater than $0.5$.

• At $n=0$, the value is $0.491525423729$
• At $n=1$, the value is $0.512186580724$
• At $n=2$, the value is $0.512186580754$
• At $n=3$, the value is $0.512186580754$

Here is a scatter diagram ,