There is another, faster, explanation, directly linked to the nature of the series: you can factor out x from each term of the series:

\(\displaystyle{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}={x}{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{{i}-{1}}}\)

\(\displaystyle={x}{\sum_{{{i}={0}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}\)

(setting \(\displaystyle{i}\to{i}-{1}\)), so

\(\displaystyle{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}={\frac{{{x}}}{{{1}-{x}}}}\)

\(\displaystyle{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}={x}{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{{i}-{1}}}\)

\(\displaystyle={x}{\sum_{{{i}={0}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}\)

(setting \(\displaystyle{i}\to{i}-{1}\)), so

\(\displaystyle{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}={\frac{{{x}}}{{{1}-{x}}}}\)