# Infinite progression - done right?

To solve: $\lim\limits_{n\to \infty}(x+x^3+x^5+...+x^{2n-1})=-\frac{2}{3}$.

So we see that it's a geometric progression (constant quotient) and it's going on infinitely. So we can apply the formula: $\sum_{n=0}^\infty a_1 q^n=\frac{a_1}{1-q}$ and we get: $\frac{x}{1-x^2}=-\frac{2}{3}$, but as it's only valid for an absolute value of the quotient less than one, we have to presume $|x^2|<1$. Solving the equation, we get $x=2$ or $x=-\frac{1}{2}$. The first is contrary to our assumption so $x=-\frac{1}{2}$.

Is this right? Or I missed something?

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The formula is only valid when $|q|\lt 1$ (not "the quotient"). Otherwise, seems fine. – Arturo Magidin Oct 16 '11 at 21:29
Looks fine to me, except that what you’re calling the quotient is actually the ratio. – Brian M. Scott Oct 16 '11 at 21:30
Thank you very much :) But $q=x^2$ so the $|x^2|<1$ still holds true, yep? – Bringiton Oct 17 '11 at 16:33

That works.

You could also note that $x+x^3+x^5+...+x^{2n-1}$ is positive when $x>0$ and does not converge when $|x| \ge 1$.

Personally I would write $x+x^3+x^5+...+x^{2n-1} = \frac{x}{1-x^2}\left(1-x^{2n}\right)$ [apart from $x=\pm 1$, when it is $nx$] before taking the limit.

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OK, thank you! :) – Bringiton Oct 17 '11 at 16:33