Evaluation of ::$S=\frac {9}{15} +\frac {19} {15^2} +\frac {119} {15^3}+\frac {1119} {15^4}\cdots $

Question

let $S$ be a sum defined as :$S=\frac {9}{15} +\frac {19} {15^2} +\frac {119} {15^3}+\frac {1119} {15^4}\cdots $, I can't succed to write this sum a geometric progression , I have a problem to deduce the closed form of $9+19+119+1119+\cdots$, Any help thanks

Answer 1

$$\sum_{n=1}^\infty\dfrac{9+\sum_{r=1}^{n-1}10^r}{15^n}$$

$$=\sum\dfrac9{15^n}+\dfrac19\left(\sum(10/15)^n-\sum(1/15)^n\right)=?$$

Answer 2

Note: $$S=\left(\frac{9}{15}+\frac{9}{15^2}+\cdots \right)+\left(\frac{10}{15^2}+\frac{100}{15^3}+\cdots \right)+\left(\frac{10}{15^3}+\frac{100}{15^4}+\cdots \right)+\cdots=$$ $$\frac{\frac{9}{15}}{1-\frac{1}{15}}+\frac{\frac{10}{15^2}}{1-\frac{10}{15}}+\frac{\frac{10}{15^3}}{1-\frac{10}{15}}+\cdots=$$ $$\frac{9}{14}+\frac{2}{15}+\frac{2}{15^2}+\cdots=\frac{9}{14}+\frac17=\frac{11}{14}.$$

Answer 3

$$S=\frac {9}{15} +\frac {19} {15^2} +\frac {119} {15^3}+\frac {1119} {15^4}\cdots$$

$$10S=\frac {90}{15} +\frac {190} {15^2} +\frac {1190} {15^3}+\frac {11190} {15^4}\cdots$$

$$\frac{10S}{15}=\frac{90}{15^2} +\frac {190} {15^3} +\frac {1190} {15^4}+\frac {11190} {15^5}\cdots$$

Subtracting we get

$$\frac{140S}{15}=\frac{90}{15} +\frac {10^2} {15^2} +\frac {10^3} {15^3}+\frac {10^4} {15^4}\cdots = \frac{90}{15} +\frac {10^2} {15^2} \frac{15}{5} = \frac{110}{15}$$

So $$S = \frac{11}{14}$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 0}^{\infty}{9 + 10\pars{10^{k} - 1}/9 \over 15^{k + 1}} & = \pars{9 - {10 \over 9}}\sum_{k = 0}^{\infty}\pars{1 \over 15}^{k + 1} + {1 \over 9}\sum_{k = 0}^{\infty}\pars{2 \over 3}^{k + 1} \\[5mm] & = {71 \over 9}\,{1/15 \over 1 - 1/15} + {1 \over 9}\,{2/3 \over 1 - 2/3} = \bbx{11 \over 14} \approx 0.7857 \end{align}