Geometrically sound

$P=\dfrac{3}{17}+\dfrac{33}{17^2}+\dfrac{333}{17^3}+ \dfrac{3333}{17^{4}}+\cdots~~(1)$ $\frac{P}{17}=\dfrac{3}{17^2}+\dfrac{33}{17^3}+\dfrac{333}{17^4}+ \dfrac{3333}{17^{5}}+\cdots~~(2)$ $\frac{16P}{17}=\dfrac{3}{17}+\dfrac{30}{17^2}+\dfrac{300}{17^3}+\cdots~~(1)-(2)$ $\frac{16P}{17}=\frac{\frac{3}{17}}{1-\frac{10}{17}}=\frac{\frac{3}{17}}{\frac{7}{17}}=\frac{3}{7}$ $P=\frac{51}{112}=\frac{m}{n}$ $\Large{\boxed{\color{#3D99F6}{m+n=163}}}$

Let the sum be $S$ , then:

$\begin{aligned} S & = \sum_{n=1}^\infty \frac{3\sum_{k=0}^n 10^k}{17^n} \\ & = \sum_{n=1}^\infty \frac{3(10^n-1)}{9\dot{}17^n} \\ & = \sum_{n=1}^\infty \frac{10^n-1}{3 \dot{} 17^n} \\ & = \frac{1}{3} \left( \sum_{n=1}^\infty \left(\frac{10}{17}\right)^n - \sum_{n=1}^\infty \left(\frac{1}{17}\right)^n \right) \\ & = \frac{1}{3} \left( \frac{1}{1-\frac{10}{17}} - \frac{1}{1-\frac{1}{17}} \right) \\ & = \frac{1}{3} \left( \frac{17}{7} - \frac{17}{16} \right) \\ & = \frac{51}{112} \\ & \\ \Rightarrow m + n & = 51 + 112 = \boxed{163} \end{aligned}$

Just let

$\displaystyle S= \dfrac{3}{17}+\dfrac{33}{17^2}+\dfrac{333}{17^3}+ \dfrac{3333}{17^{4}}+\cdots$ $(1)$

$\displaystyle \Longrightarrow 17S = 3 +\dfrac{33}{17}+\dfrac{333}{17^2}+ \dfrac{3333}{17^{3}}+\cdots$ $(2)$

Now, $(2) - (1)$ gives

$\displaystyle 16S= 3 +\dfrac{30}{17}+\dfrac{300}{17^2}+ \dfrac{3000}{17^{3}}+\cdots \Longleftrightarrow 16S = 3 + \dfrac{30}{7} = \dfrac{51}{7}$

Thus, $\displaystyle S = \dfrac{51}{112}$ and $\displaystyle m + n = 51 + 112 = 163$ .

I split the numerators into 3, 30, 300, and so on. We must find $3\cdot\left(\frac1{17} + \frac1{17^2} + \cdots\right) + 30\cdot\left(\frac1{17^2} + \frac1{17^3} + \cdots\right) + \cdots \\ = \left(3 + 30\cdot \frac1{17} + 300\cdot \frac1{17^2} + \cdots\right)\cdot\left(\frac1{17} + \frac1{17^2} + \cdots\right) \\ = 3\cdot \left(1 + \frac{10}{17} + \frac{10^2}{17^2} + \cdots\right)\cdot\left(\frac1{17} + \frac1{17^2} + \cdots\right) \\ = 3\cdot \frac{17}{7}\cdot \frac{1}{16} = \frac{3\cdot 17}{7\cdot 16} = \frac{51}{112},$ giving the answer $\boxed{163}$ .

$10^n - 1 = 9999. . . 9, n ~9s. \\ \therefore \dfrac{3}{17} + \dfrac{33}{17^2} + \dfrac{333}{17^3} . . . \\ \frac 1 3 *\left \{ \dfrac{10}{17} - \dfrac{1}{17} +\dfrac{10^2}{17^2} - \dfrac{1}{17^2} + \dfrac{10^3}{17^3} - \dfrac{1}{17^3} . . . \right \}\\ = \frac 1 3 *\left \{ \dfrac{10}{17} +\dfrac{10^2}{17^2} + \dfrac{10^3}{17^3} . . .~~~~~ - \dfrac{1}{17} - \dfrac{1}{17^2} - \dfrac{1}{17^3} . . .\right \}\\ =\frac 1 3 *\left \{ \dfrac 1 {1 - \frac{10}{17} } ~~ - \dfrac1 {1 - \frac{1}{17} } \right \} \\ =\dfrac {51}{112}=\dfrac {m}{n}. ~~m+n=163.$