Maclaurin series expansion of $\frac{1}{(1+x)^n}$

Question

I am trying to figure out the Maclaurin Series expansion of the function, preferribly in a sneaky and clever way. Any ideas?

Thanks.

Answer 1

It's almost the $n^{\rm th}$ derivative of $1/(1+x)$.

Answer 2

set $$ F(x)=\frac1{(1-x)^n} = \prod_{j=1}^n\sum_{k=0}^{\infty}x^k $$ here the coefficient of $x^m$ is the number of ways of expressing $m$ as a sum of $n$ non-negative integers, with order significant (i.e. $3+4+5$ and $4+3+5$ are different expressions).

this is just the number of ways of fitting $n-1$ separators into $m$ slots, i.e. $\binom{n+m-1}{m}$ so: $$ \frac1{(1+x)^n}=F(-x)=1-nx+\frac{n(n+1)}{2!}x^2-... $$

Answer 3

I think that computing the derivatives is really the easiest way (as pointed out By Andre Nicolas). Your function is

$$ f(x) = (1+x)^{-n}\,. $$

Now differentiate: \begin{eqnarray} f'(x) &=& -n (1+x)^{-(n+1)} \\ f''(x) &=& (-1)^2 n(n+1) (1+x)^{-(n+2)}\,. \end{eqnarray} By now the general pattern should be clear:

$$ f^{(k)} (x) = (-1)^k \, n (n+1) \cdots (n+k-1) \, (1+x)^{-n+k}. $$

Evaluating the above at $x=0$ you get the following Maclaurin series (convergent for $|x|<1$)

$$ f(x) = \sum_{k=0}^{\infty} a_k\, x^k$$

with

$$ a_k = \frac{f^{(k)}(0)}{k!} = (-1)^k \frac{ n (n+1) \cdots (n+k-1)}{k!} = (-1)^k \left ( \begin{array}{c} n+k-1\\ k \end{array} \right ). $$

Answer 4

Simple, yet not straightforward, approach: \begin{align} \frac{1}{(1-x)^n} &= (1-x)^{-n} \\ &= \sum_{k=0}^\infty\binom{-n}{k}(-x)^k\tag 1 \\ &= \sum_{k=0}^\infty\left((-1)^k\binom{n+k-1}{k}\right)(-x)^k\tag2 \\ &= \sum_{k=0}^\infty\binom{n+k-1}{k}x^k \\ \end{align}

We arrive at $(1)$ using the binomial series, and at $(2)$ using the relationship between the multiset coefficient and the binomial coefficient $\binom{-n}{k}$.