I might be inclined to go something along these lines...

$\displaystyle \begin{align*} \frac{a^m}{ \left( 1 + a \right) ^m } &= \left( \frac{a}{1 + a} \right) ^m \\ &= \left( \frac{1 + a - 1 }{1 + a} \right) ^m \\ &= \left( 1 - \frac{1}{1 + a} \right) ^m \\ &= \left[ 1 - \frac{1}{1 - \left( - a \right) } \right] ^m \\ &= \left\{ 1 - \sum_{n = 0}^{\infty}{ \left[ \left( -a \right) ^n \right] } \right\} ^m \\ &= \left\{ - \sum_{n = 1}^{\infty} \left[ \left( -a \right) ^n \right] \right\} ^m \end{align*}$

And now you can probably get a decent approximation using a binomial expansion. Of course this is only convergent where |a| < 1.