Logariddim

We rewrite the infinite sum in sigma form:

$\sum_{n=0}^{\infty} (\log_{b} 2)^{n}(\log_{b} 5^{2^{n}})$

Using the properties of logarithms and the formula of infinite GP, we obtain

$\sum_{n=0}^{\infty} (\log_{b} 2)^{n}(\log_{b} 5^{2^{n}}) = \log_{b} 5 \sum_{n=0}^{\infty} (2\log_{b} 2)^{n} = \frac{\log_{b} 5}{1-2\log_{b} 2} = \frac{\log_{b} 5}{2\log_{b} 5} = \boxed{0.5}$

$\text{Let } S \text{ be the infinite sum}$

$S=\displaystyle\sum_{n=0}^\infty{(\log_{100} 2)}^n(\log_{100} {5^{2^n}})$

$S=\displaystyle\sum_{n=0}^\infty{(\log_{100} 2)}^n(2^n\log_{100} 5)$

$S=(\log_{100} 5) \displaystyle\sum_{n=0}^\infty{2^n(\log_{100} 2)}^n$

$S=(\log_{100} 5) \displaystyle\sum_{n=0}^\infty{(2\log_{100} 2)}^n$

$S=(\log_{100} 5) \displaystyle\sum_{n=0}^\infty{(\log_{100} 4)}^n$

$\displaystyle\sum_{n=0}^\infty{(\log_{100} 4)}^n \text{ is geometric } \\ \text{with }r=\log_{100} 4$

$S=\log_{100} 5\frac{1}{1-\log_{100} 4}$

$S=\frac{\ln 5}{\ln {100}} \cdot \frac{1}{1-\frac{\ln 4}{\ln {100}}}$

$S=\frac{\ln 5}{\ln {100} - \ln 4}$

$S=\frac{\ln 5}{\ln {25}}$

$S=\frac{1}{\frac{\ln {25}}{\ln 5}}$

$S=\frac{1}{\log_5 {25}}$

$\boxed{S=\frac{1}{2}}$