Too many 5s

$\begin{aligned} S & = \frac{5}{13} + \frac{55}{13^2} + \frac{555}{13^3} + \ldots \\ \frac{S}{13} & = \frac{5}{13^2}+\frac{55}{13^3}+\frac{555}{13^4}+\ldots \\ S-\frac{S}{13} & = \frac{12S}{13} = \frac{5}{13}+\frac{50}{13^2}+\frac{500}{13^3}+\ldots \\ & = \frac{5}{13}+\frac{ \frac{50}{13^2} }{1-\frac{10}{13}} = \frac{5}{13}+\frac{50}{13×3}=\frac{5}{3} \\ S & = \frac{13}{12} \cdot \frac{5}{3}=\frac{65}{36} \Rightarrow 65+36=\boxed{101} \end{aligned}$

Begin by taking out the common factor 5 to get

$S=5\left(\frac{1}{13}+\frac{11}{13^2}+\frac{111}{13^3}\dots\right)$

Now write

$111\dots111$ with n ones as $\frac{10^n - 1}{9}$

(e.g. $\frac{10^3 - 1}{9}=\frac{999}{9}=111$ )

and take out the common factor $\frac{1}{9}$ to write the sum as

$S=\frac{5}{9}\left(\sum_{n=1}^{\infty}\left(\frac{10^n - 1}{13^n}\right)\right)$

Separate the summand into two geometric series to get

$S=\frac{5}{9}\left(\sum_{n=1}^{\infty}\left(\left(\frac{10}{13}\right)^n-\left(\frac{1}{13}\right)^n\right)\right)$

Now use the formula for summing infinite geometric series to get

$S=\frac{5}{9}\left(\frac{\frac{10}{13}}{1-\frac{10}{13}}-\frac{\frac{1}{13}}{1-\frac{1}{13}}\right)$

and so

$S=\frac{5}{9}\left(\frac{10}{3}-\frac{1}{12}\right)=\frac{5}{9}\frac{39}{12}=\frac{65}{36}=\frac{a}{b}$

and so

$a+b=65+36=\boxed{101}$

$S = 5\cdot \left(\frac 1{13} + \frac 1{13^2} + \cdots\right) + 50\cdot \left(\frac 1{13^2} + \frac 1{13^3} + \cdots\right) + \cdots \\ = 5\cdot \left(1 + \frac{10}{13} + \frac{10^2}{13^2} + \cdots\right)\cdot \left(\frac 1{13} + \frac 1{13^2} + \cdots\right) \\ = 5\cdot \frac{1}{1-10/13} \cdot \frac{1/13}{1-1/13} = 5\cdot \frac{13}{3} \cdot \frac{1}{12} = \frac{65}{36},$ so that the answer is $65 + 36 = \boxed{101}$ .

Exactly same here too. Interesting series.