An algebra problem by أحمد الحلاق

Let the sum be $S$ , then we have

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {\log_{10} 2^{2n-1}}{2\cdot 5^{n-1}} \\ & = \sum_{n=\color{#D61F06}{0}}^\infty \frac {\log_{10} 2^{\color{#D61F06} {2n+1}}}{2\cdot 5^{\color{#D61F06}{n}}} \\ & = \sum_{n=0}^\infty \frac {(2n+1)\log_{10} 2}{2\cdot 5^n} \\ & = \frac {\log_{10}2}2 \sum_{n=0}^\infty \frac {2n+1}{5^n} \\ & = \frac {\log_{10}2}2 \left(2 \color{#3D99F6}{\sum_{n=0}^\infty \frac n{5^n}} + \sum_{n=0}^\infty \frac 1{5^n} \right) & \small \color{#3D99F6}{\text{See Note.}} \\ & = \frac {\log_{10}2}2 \left(2 \times \color{#3D99F6}{\frac {\frac 15}{\left(1-\frac 15\right)^2}} + \frac 1{1-\frac 15} \right) \\ & = \frac {\log_{10}2}2 \left(\frac 58 + \frac 54 \right) \approx \boxed{0.282} \end{aligned}$

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Note:

See arithmetic-geometric progression (AGP) or as follows.

$\begin{aligned} \frac x{1-x} & = \sum_{n=1}^\infty x^n & \small \color{#3D99F6}{\text{Differentiate both sides w.r.t. }x} \\ \frac 1{(1-x)^2} & = \sum_{n=1}^\infty nx^{n-1} & \small \color{#3D99F6}{\text{Multiply both sides by }x} \\ \frac x{(1-x)^2} & = \sum_{n=1}^\infty nx^n & \small \color{#3D99F6}{\text{put }x = \frac 15} \\ \frac {\frac 15}{\left(1-\frac 15 \right)^2} & = \sum_{n=1}^\infty \frac n{5^n} \end{aligned}$

Let S = $\cfrac{log 2}{2} + \cfrac{3log 2}{2*5} + \cfrac{5log2}{2*5^{2}} + \cfrac{7log2}{2*5^{3}} + ...$

=> 2S = $log2 + \cfrac{3log2}{5} + \cfrac{5log2}{5^{2}} + \cfrac{7log2}{5^{3}} + ...$

Let V = $\cfrac{3log2}{5} + \cfrac{5log2}{5^{2}} + \cfrac{7log2}{5^{3}} + ...$

=> 5V = $3log2 + \cfrac{5log2}{5} + \cfrac{7log2}{5^{2}} + ...$

=> 4V = $3log2 + \cfrac{2log2}{5} + \cfrac{2log2}{5^{2}} + ...$

=> V = $\cfrac{1}{4} * (3log2 + \cfrac{\cfrac{2log2}{5}}{1 - \cfrac{2log2}{5}})$

=> S = $\cfrac{log2}{2} + \cfrac{1}{8}(3log2 + \cfrac{2log2}{5-2log2})$ ~ $0.281$