Series approximation to $\frac{1}{1-(1+x)^{-n}}$

Question

I'm attempting to find a series approximation to:

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

where $n\in\mathbb{N^+}$ and is a constant, and $x\in\mathbb{R}$ and $0<x<1$.

Using Wolfram Alpha, I noted the first order approximation is:

$$f(x)\approx\frac{1}{nx} + \frac{n+1}{2n}$$

With this goal post in mind, I went about trying to find the first order approximation for myself. The original formula has a singularity at $x=0$, so I figured I can't use the Taylor series formula directly. Instead, with the geometric series formula I found the following:

$$f(x)=\sum_{i=0}^\infty\frac{1}{(1+x)^{in}} $$

But I'm not sure how to proceed from here. Is there a trick I'm missing?

Looking around for a hint on how to proceed, I noticed that $f(x)$ has some similarities to mortgage monthly payments formula:

$$C(r)=\frac{r(1+r)^nP}{(1+r)^n-1}$$

where $P$ is the initial load size (principal), $r$ is the interest rate per month and $n$ is the number of payments (months). I found similar questions here and here, but it's still not clear to me what the initial steps are for finding the first order approximation given by Wolfram Alpha.

Any help would be much appreciated, even if it's just the first few steps towards the solution. Thanks.

Answer 1

For any positive integer $n$, $$\eqalign{f(x) &= \frac{1}{1 - (1+x)^{-n}} = \frac{(1+x)^n}{(1+x)^n - 1}\cr &= \frac{(1+x)^n}{\displaystyle\prod_{j=0}^{n-1} ((x + 1 - e^{2\pi i j/n}) }}$$ This is a rational function with limit $1$ as $x \to \infty$ and simple poles at $e^{2\pi i j/n}-1$ for $j = 0$ to $n-1$, i.e. $\omega- 1$ for the $n$'th roots of unity $\omega$. Thus it can be written as $$ \eqalign{f(x) &= 1 + \sum_{j=0}^{n-1} \frac{c_j}{x + 1 - e^{2\pi i j/n}}\cr &= 1 + \frac{c_0}{x}- \sum_{j=1}^{n-1} c_j \sum_{k=0}^\infty (-1+e^{2\pi i j/n})^{-1-k} x^k}$$ for appropriate constants $c_j$.

Answer 2

Start by multiplying both numerator and denominator by $(1+x)^n$: $$f(x)=\frac{1}{1-(1+x)^{-n}}=\frac{(1+x)^n}{(1+x)^n-1}$$ Now let's use binomial expansion $$(1+x)^n=1+nx+\frac{n(n-1)}2x^2+...$$ This yields $$f(x)=\frac{1+nx+\frac{n(n-1)}2x^2+...}{nx+\frac{n(n-1)}2x^2+...}\\=\frac1{nx}\frac{1+nx+\frac{n(n-1)}2x^2+...}{1+\frac{n-1}2x+...}$$ We can now use $$\frac1{1+\frac{n-1}2x+...}\approx1-\frac{n-1}2x$$ This yields $$f(x)\approx\frac1{nx}\left(1+nx+\frac{n(n-1)}2x^2+...\right)\left(1-\frac{n-1}2x\right)\\\approx\frac1{nx}\left(1+nx-\frac{n-1}2x\right)\\=\frac1{nx}\left(1+\frac{n+1}2x\right)=\frac1{nx}+\frac{n+1}{2n}$$ Note that at my last approximation I've dropped terms in $x^2$. If I want to keep those, I need one more term in an earlier expression.