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$\text{Find the general sum formula for: } \sum_{n=1}^{\infty}x^{2n-1}, x\in(-1,1)$. I know how to go about this if it were just $\sum_{n=1}^{\infty}x^{n-1}$, as it is just $\frac{1}{1-x}$, however the exponent of $2n-1$ is where I get confused. Any help would be greatly appreciated.

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    $\begingroup$ By the way, it could be worth knowing the general geometric series formula: $$a+ ar+ ar^2 + \cdots = \frac{a}{1-r}$$ (for $|r|<1$), where $a$ is the first term in the series, and $r$ is the common ratio. There is also a formula like this for a finite geometric series: $$a+ar+ar^2+\cdots+ar^{n-1} = \frac{a\left(1-r^n\right)}{1-r},$$ for $r\ne 1$. Here $a$ is the starting term, $r$ is the common ratio, and $n$ is the number of terms in the series. $\endgroup$ – Minus One-Twelfth Mar 21 '19 at 1:57
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Hint: Note that $$\sum_{n=1}^{\infty}x^{2n-1} = x + x^3 + x^5 + \cdots = x(1+ x^2 + x^4+\cdots).$$ Do you know how to sum the expression in the brackets? If not, try substituting $u=x^2$ and note that that series becomes $1+u+u^2+\cdots$.

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$$\sum_{n=1}^{\infty}x^{2n-1}=\sum_{n=0}^{\infty}x^{2n+1} = x \sum_{n=0}^{\infty}x^{2n} = x \sum_{n=0}^{\infty}(x^2)^n=\frac{x}{1-x^2}.$$

Note the change in the limits of summation.

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