# I have to solve - \sum_{i=1}^\infty(\frac{5}{12})^i - geometric series?

I have to solve -
$$\displaystyle{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}$$
- geometric series?

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rodclassique4r
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
$$\displaystyle{\sum_{{{i}={1}}}^{\infty}}{\left({\frac{{{5}}}{{{12}}}}\right)}^{{i}}={\frac{{{x}}}{{{1}-{x}}}}$$
###### Not exactly what youâ€™re looking for?
Shannon Hodgkinson
HINT:
$$\displaystyle{\sum_{{{i}={0}}}^{\infty}}{x}_{{i}}={\frac{{{1}}}{{{1}-{x}}}}={x}_{{0}}+{\sum_{{{i}={1}}}^{\infty}}{x}_{{i}}$$
karton

Observe that for suitable x:
$$(1-x)\sum_{i=k}^\infty x^i=\sum_{i=k}^\infty x^i-\sum_{i=k+1}^\infty x^i=x^k$$
so that:
$$\sum_{i=k}^\infty x^i=\frac{x^k}{1-x}$$
It remains to substitute k=1 and $$x=\frac{5}{12}$$