I have to solve - ∑i=1∞(512)i - geometric series?

HINT: ∑i=0∞xi=11−x=x0+∑i=1∞xi

Observe that for suitable x: (1−x)∑i=k∞xi=∑i=k∞xi−∑i=k+1∞xi=xk so that: ∑i=k∞xi=xk1−x It remains to substitute k=1 and x=512

HINT: ∑i=0∞xi=11−x=x0+∑i=1∞xi

Observe that for suitable x: (1−x)∑i=k∞xi=∑i=k∞xi−∑i=k+1∞xi=xk so that: ∑i=k∞xi=xk1−x It remains to substitute k=1 and x=512

Answered: "I have to solve - ∑i=1∞(512)i - geometric..."

Talamancoeb

Answered question

2022-01-07

I have to solve -

\sum_{i = 1}^{\infty} {(\frac{5}{12})}^{i}

- geometric series?

Answer & Explanation

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

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

(setting

i \to i - 1

), so

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

Shannon Hodgkinson

Beginner2022-01-09Added 34 answers

HINT:

\sum_{i = 0}^{\infty} x_{i} = \frac{1}{1 - x} = x_{0} + \sum_{i = 1}^{\infty} x_{i}

karton

Expert2022-01-11Added 613 answers

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 $x = \frac{5}{12}$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?