Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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


share|improve this question
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
add comment

1 Answer

$\displaystyle \sum \frac{1}{(1+x^b)^n} = \frac{1}{1-\frac{1}{(1 + x^b)}} = \frac{1 + x^b}{x^b} \implies \sum \frac{x^a}{(1 + x^b)^n} = \frac{x^a + x^{b+a}}{x^b} = x^{a-b} + x^b$

where this is only valid for $x > 0$ as we used geometric series expansions on a ratio of $\frac{1}{1+x^b}$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.