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Value of $\sum\limits_n x^n$

If I have some real $x$ where $0 < x < 1$

What is the value $y = x + x^2 + x^3 + x^4 + \dots$ ?

Intuitively I can see that for $x = 0.5$ then $y = 1$

How do I calculate this for arbitrary $x$?

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marked as duplicate by Pedro Tamaroff, Aryabhata, J. M., t.b., Asaf Karagila Apr 27 '12 at 11:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

This is a geometric series. – Adrián Barquero Apr 15 '12 at 20:13
What you've written, $0>x>1$, makes no sense. Presumably you mean to write $0<x<1$? – Keenan Kidwell Apr 15 '12 at 20:14
@KeenanKidwell: It makes perfect sense, it's just always false. :) I've corrected the type, thanks. – Andrew Tomazos Apr 15 '12 at 20:52
up vote 1 down vote accepted

(You mean $0<x<1$.)

This is just a geometric series with first term and ratio $x$, so $$y=\frac{x}{1-x}\;.$$

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If you don't know much about series maybe the following is helpful.

Suppose that such a sum exists. It is clear that $y=x+x(x+x+x^2+\cdots)=x+xy.$ Just find $y$ from the equation $y=x+xy.$ (I'm neglecting some limits here).

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Do not memorize the formula. You can derive it using the following trick. Let $s=x+x^2+x^3+...$. Then you have that $sx=x^2+x^3+x^4+...$. Hence, $s-sx=x$. In other words, $s=\frac{x}{1-x}$

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Some find remembering the formula superior to re-deriving it every time. Others not. – GEdgar Apr 27 '12 at 3:05
Actually my series was $y = 6x - 2x^2 + 6x^3 - 2x^2 + 6x^5 + ...$, so the derivation helped much more than the formulae – Andrew Tomazos Apr 27 '12 at 7:18

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