Is it really a GP?

$\begin{aligned} S & = \frac 3{13} + \frac {33}{13^2} + \frac {333}{13^3} + \frac {3333}{13^4} + \cdots \\ & = \frac 9{3 \cdot 13} + \frac {99}{3 \cdot 13^2} + \frac {999}{3 \cdot 13^3} + \frac {9999}{3 \cdot 13^4} + \cdots \\ & = \frac {10-1}{3 \cdot 13} + \frac {100-1}{3 \cdot 13^2} + \frac {1000-1}{3 \cdot 13^3} + \frac {10000-1}{3 \cdot 13^4} + \cdots \\ & = \sum_{n=1}^\infty \frac {10^n-1}{3 \cdot 13^n} \\ & = \frac 13 \sum_{n=1}^\infty \left[ \left(\frac {10}{13}\right)^n - \frac 1{13^n} \right] \\ & = \frac 13 \left[ \frac {\frac {10}{13}}{1-\frac {10}{13}} - \frac {\frac 1{13}}{1-\frac 1{13}} \right] \\ & = \frac 13 \left[ \frac {10}3 - \frac 1{12} \right] \\ & = \boxed{\dfrac {13}{12}} \end{aligned}$

$S=\frac{3}{13}+\frac{33}{13^2}+\frac{333}{13^3}+\frac{3333}{13^4}+\ldots$

$\frac{S}{13}=\frac{3}{13^2}+\frac{33}{13^3}+\frac{333}{13^4}+\ldots$

$S-\frac{S}{13}=\frac{3}{13}+\frac{30}{13^2}+\frac{300}{13^3}+\frac{3000}{13^4}$

$=\frac{\frac{3}{13}}{1-\frac{10}{13}}=1$

$S-\frac{S}{13}=\frac{12S}{13}=1$

$\boxed{S=\frac{13}{12}}$

See the denominator is a G.P. but what is upper one just increase digit. - = $\frac{3}{13}$ + $\frac{33}{13^2}$ + $\frac{333}{13^3}$ . . . . ∞

• =3( $\frac{1}{13}$ + $\frac{11}{13^2}$ + $\frac{111}{13^3}$ . . . . .∞)

• = $\frac{3}{9}$ ( $\frac{9}{13}$ + $\frac{99}{13^2}$ + $\frac{999}{13^3}$ . . . . . ∞)

• = $\frac{3}{9}$ ( $\frac{10-1}{13}$ + $\frac{100-1}{13^2}$ + $\frac{1000-1}{13^3}$ . . . . . ∞)

• = $\frac{3}{9}$ [( $\frac{10}{13}$ + $\frac{100}{13^2}$ + $\frac{1000}{13^3}$ ). . . . . ∞( $\frac{3}{9}$ ( $\frac{-1}{13}$ + $\frac{-1}{13^2}$ + $\frac{-1}{13^3}$ . . . . . ∞)]

• = $\frac{3}{9}$ [( $\frac{10\13}{1-10/13}$ ) + ( $\frac{-1/13}{1-1/13}$ )]

• = $\frac{1}{3}$ [ $\frac{10}{3}$ + $\frac{-1}{12}$ ]

• = $\frac{1}{3}$ $\frac{39}{12}$

• = $\boxed{13/12}$