Series Expansion of $\frac{1}{(x+a)^n}$ Is there a simple series expansion for:
$$ \frac{1}{(x+a)^n}$$
Where $|a| &lt; x$ and $x \in \Bbb R^+$

I am interested in a one that does not contain the $(x+a)^n$.
If $|x| &lt; |a|$,
$\frac1{(a+x)^n}
=\frac1{a^n(1+x/a)^n}
=\frac1{a^n}(1+x/a)^{-n}
=\frac1{a^n}\sum_{k=0}^{\infty} \binom{-n}{k}\left(\frac{x}{a} \right)^k
$

If $|x| &gt; |a|$,
$\frac1{(a+x)^n}
=\frac1{x^n(1+a/x)^n}
=\frac1{x^n}(1+a/x)^{-n}
=\frac1{x^n}\sum_{k=0}^{\infty} 
\binom{-n}{k}\left(\frac{a}{x} \right)^k
=\sum_{k=0}^{\infty} 
\binom{-n}{k}\frac{a^k}{x^{n+k}}
$

If you want, you can apply
$\binom{-n}{k}
=(-1)^k\binom{n+k-1}{k}
$.
HINT:

Write it as

$$(x + a)^{-n}$$

and use Binomial expansion.

More

https://en.wikipedia.org/wiki/Binomial_theorem

To understand the two cases in which $|x| &gt; |a|$ and $|x| &lt; |a|$ just read the answer above!