Possible Duplicate:
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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asked Apr 15 '12 at 20:12
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$\begingroup$ This is a geometric series. $\endgroup$ – Adrián Barquero Apr 15 '12 at 20:13
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$\begingroup$ What you've written, $0>x>1$, makes no sense. Presumably you mean to write $0<x<1$? $\endgroup$ – Keenan Kidwell Apr 15 '12 at 20:14
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$\begingroup$ @KeenanKidwell: It makes perfect sense, it's just always false. :) I've corrected the type, thanks. $\endgroup$ – Andrew Tomazos Apr 15 '12 at 20:52
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(You mean $0<x<1$.)
This is just a geometric series with first term and ratio $x$, so $$y=\frac{x}{1-x}\;.$$
answered Apr 15 '12 at 20:14
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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, $s=\frac{x}{1-x}$
answered Apr 27 '12 at 0:07
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$\begingroup$ Some find remembering the formula superior to re-deriving it every time. Others not. $\endgroup$ – GEdgar Apr 27 '12 at 3:05
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$\begingroup$ Actually my series was $y = 6x - 2x^2 + 6x^3 - 2x^2 + 6x^5 + ...$, so the derivation helped much more than the formulae $\endgroup$ – Andrew Tomazos Apr 27 '12 at 7:18
