Talamancoeb

2022-01-07

I have to solve -

$\sum _{i=1}^{\mathrm{\infty}}{\left(\frac{5}{12}\right)}^{i}$

- geometric series?

- geometric series?

rodclassique4r

Beginner2022-01-08Added 37 answers

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

$\sum _{i=1}^{\mathrm{\infty}}{\left(\frac{5}{12}\right)}^{i}=\frac{x}{1-x}$

(setting

Shannon Hodgkinson

Beginner2022-01-09Added 34 answers

HINT:

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

karton

Expert2022-01-11Added 613 answers

Observe that for suitable x:

so that:

It remains to substitute k=1 and

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