Talamancoeb

2022-01-07

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

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:
$\sum _{i=1}^{\mathrm{\infty }}{\left(\frac{5}{12}\right)}^{i}=x\sum _{i=1}^{\mathrm{\infty }}{\left(\frac{5}{12}\right)}^{i-1}$
$=x\sum _{i=0}^{\mathrm{\infty }}{\left(\frac{5}{12}\right)}^{i}$
(setting $i\to i-1$), so
$\sum _{i=1}^{\mathrm{\infty }}{\left(\frac{5}{12}\right)}^{i}=\frac{x}{1-x}$

Shannon Hodgkinson

HINT:
$\sum _{i=0}^{\mathrm{\infty }}{x}_{i}=\frac{1}{1-x}={x}_{0}+\sum _{i=1}^{\mathrm{\infty }}{x}_{i}$

karton

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

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