# Show that the following infinite series of x is equal to the given function.

Here's the question:

$$\sum_{n=0}^{\infty}\frac{x^{\frac{3}{2}}}{\left ( 1+x^{2} \right )^{n}}= \left\{\begin{matrix} 0 &\left (x=0 \right ) \\ x^{\frac{-1}{2}}+x^{\frac{3}{2}} & \left ( 0< x\leq 1 \right ) \end{matrix}\right.$$

Show that this is true. (I'd be glad if the approach is constructive, instead of backtracking by Taylor series). And also, can we find a general formula for

$$\sum_{n=0}^{\infty}\frac{x^\alpha }{\left ( 1+x^{\beta } \right )^{n}}$$

Thanks.

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Oh, I didn't realize that this was a simple power series. Sorry about the question. – firemind Mar 7 '12 at 1:51
Simple geometric series may be what you mean. – Gerry Myerson Mar 7 '12 at 1:52
@Gerry Myerson yes :) – firemind Mar 7 '12 at 2:38
@firemind $$\sum\limits_{n \geqslant 0} {\frac{{{x^a}}}{{{{\left( {1 + {x^b}} \right)}^n}}}} = {x^{a - b}} + {x^a}$$ – Pedro Tamaroff Mar 7 '12 at 2:44

$\displaystyle \sum \frac{1}{(1+x^b)^n} = \frac{1}{1-\frac{1}{(1 + x^b)}} = \frac{1 + x^b}{x^b} \implies \sum \frac{x^a}{(1 + x^b)^n} = \frac{x^a + x^{b+a}}{x^b} = x^{a-b} + x^b$
where this is only valid for $x > 0$ as we used geometric series expansions on a ratio of $\frac{1}{1+x^b}$