I solved this equation something like this, as shown in the photo:
Is it correct?
If I put $x=2$ I get weird results!
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Sad news for you: that works only with an infinite number of terms, and for $|x|<1$. – Intelligenti pauca Nov 16 '15 at 14:55
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As long as you wrote $y=x+x^2+x^3+ . . . $ , thus you suppose your series is convergent with sum equals $y$. However this is a geometric series $\sum_{k=1}^\infty x^k$, which is convergent for $|x| < 1$ – Nizar Nov 16 '15 at 14:56
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the sum of geometric series is given by $a(r^n - 1/ (r-1)$ here $r=x , a=x$ – Archis Welankar Nov 16 '15 at 14:57
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I am not surprised that you got strange results for $x=2$, since you're in that case trying to calculate $2+4+8+16+\cdots$, which isn't really possible (at least not conventionally). Perhaps $x=1/2$ will give you more consistent results. – Arthur Nov 16 '15 at 14:57
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remember $mod x$ should be less than $1$ for a series to be convergent. – Archis Welankar Nov 16 '15 at 14:59
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It's true that
$$x + x^2 + x^3 + ... = \frac{x}{1-x} $$
provided that $|x|< 1$. Therefore you cannot apply it when $x=2$.
In short, the reason it does not work when $|x|>1$ is because the terms blow up (the series is said to be divergent).
Eff
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How do you know it doesn’t work for x>1? Is it a kind of theorem? – Aaryan Dewan Nov 16 '15 at 15:21
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The reason is, when we have an infinite series (adding an infinite number of terms), it is required that the terms tend to zero, see here. When $|x|>1$, for example if $x=2$, the terms $x, x^2, x^3, ...$ get larger and larger and do not add up to a finite number. – Eff Nov 16 '15 at 15:29
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But if you go on adding numbers smaller than 0 , won’t they add up to infinite? – Aaryan Dewan Nov 16 '15 at 15:48
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As I said, the terms have to tend to zero. For example if $x = -1/2$, then the sequence $x, x^2, x^3,...$ is $$-\frac12, \frac14, -\frac18, ... \to 0 $$ and the series $$\sum\limits_{i=1}^\infty \left(-\frac12\right)^i$$ does in fact converge. With sum $$\frac{x}{1-x} = \frac{-1/2}{1+1/2} = -\frac13. $$ – Eff Nov 16 '15 at 15:56
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You assumed in your proof that the quantity $x + x^2 + x^3 + \cdots$ is a number. Indeed, whenever $(x + x^2 + x^3 + \cdots)$ is a quantity satisfying the formal rules of arithmetic, we must have $$ x + x^2 + x^3 + \cdots = \frac{x}{1-x} $$ as you have proven. However, under the usual considerations, $x + x^2 + x^3 + \cdots$ is not a real number. For example, if $x > 1$, it is more typical to say something like $$ x + x^2 + x^3 + \cdots = \infty $$ and $\infty$ does not satisfy the formal rules of arithmetic (for example, $\infty - 1 = \infty$).
It turns out that your formula only makes sense in our usual understanding of infinite sums when $|x| < 1$. Perhaps you can see how $x + x^2 + x^3 + \cdots$ should only make sense if the terms $x,x^2,x^3,\dots$ get smaller and smaller.
Ben Grossmann
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As a lot of the comments and the other answers point out, this only works for $-1<x<1$. I want to go in a slightly different direction: If you want to make sense of the function $\frac{x}{1-x}$ as a geometric series outside of that interval, we have: $$ -1 - \frac 1x - \frac{1}{x^2} - \frac{1}{x^3} - \cdots $$ which is a sequence that converges as long as $|x|> 1$, and it gives $\frac{x}{1-x}$ (or really, $\frac{1}{1/x-1}$, which ammounts to the same thing) if you do the same trick as you've done in your question. This is what is called "the series expansion of $\frac x{1-x}$ around $\infty$" (since the series converges as long as $x$ is large enough), while $x + x^2 + x^3 + \cdots$ is the series expansion around $0$ (since it converges as long as $x$ is close enough to $0$ [there is a bit more to it than that, but that's details]). Together, they give you geometric series that evaluate to $\frac{x}{x-1}$ on the whole number line except at $-1$ and $1$.
Arthur
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Can you please explain me what is the meaning of converges as long as |x| > 1 ? – Aaryan Dewan Nov 16 '15 at 15:17
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It meas that if $x$ is a number so that $|x|>1$, then the series $-1-1/x - 1/x^2 - \cdots$ is convergent, and converges to $\frac{x}{1-x}$. – Arthur Nov 16 '15 at 15:21