prove that $ 1 + (1+a)x + (1+a+a^{2})x^{2}$ + ....$\infty$ is equal to $\frac{1}{(1-x)(1-ax)}$

Question

Question

If $a$,$ x $are real numbers and $|a| < 1$, $|x| < 1$, then prove that $ 1 + (1+a)x + (1+a+a^{2})x^{2}$ + ....$\infty$ is equal to $\frac{1}{(1-x)(1-ax)}$

Hint:

$\frac{1}{1-x}+\frac{ax}{1-x}+\frac{a^2 x^2}{1-x}+...=\frac{1}{1-x}.(1+ax+a^2 x^2+...)=\frac{1}{1-x}.\frac{1}{1-ax}$

I can't understand how i would get the step given in the hint from the question since $\frac{1}{1-x}$ is not given in the question but is given in the hint.Except this step i can solve the problem on my own.Any help would be appreciated.

Answer 1

$$S_n = \frac{1-a}{1-a}+ \frac{1-a^2}{1-a}x + \cdots + \frac{1-a^{n+1}}{1-a}x^n \rightarrow$$ $$(1-a)S_n = \sum_{k=0}^nx^k - a\sum_{k=0}^n(ax)^k \rightarrow$$ $$S_n \stackrel{n\rightarrow \infty}{\rightarrow}S = \frac{1}{1-a}(\frac{1}{1-x} - \frac{a}{1-ax})= \frac{1}{(1-x)(1-ax)}$$

Answer 2

EDIT: Replacing "Taylor series" with "geometric series" still keeps everything true, except now it is more easy to understand (summing geometric series is an easy proof).

Do you know how to compute the Taylor series of $\frac{1}{1-x}$ and $\frac{1}{1-ax}$?

Notice that $(1+x+x^2+x^3+\cdots)(1+ax+a^2x^2+a^3x^3+\cdots)$ is equal to the sum given in the beginning. (Try and see why this is true for yourself—it's easy.) For $\lvert x\rvert <1$, the left sum is the Taylor series for $\frac{1}{1-x}$, so that gives you the first quantity in the hint. After that, just follow the equalities in the hint.

Also, the sum on the right is the series for $\frac{1}{1-ax}$, given the appropriate bounds on $a$ and $x$. This should immediately prove the problem if you believe that you can multiply Taylor series within their radii of convergence (i.e. you understand why this is true and can prove it perhaps).

Answer 3

Method 1: Expand the brackets and regroup: $$(1+x+x^2+\cdots)+(ax+ax^2+ax^3+\cdots)+(a^2x^2+a^2x^3+a^2x^4+\cdots)+\cdots=$$ $$\frac{1}{1-x}+\frac{ax}{1-x}+\frac{a^2x^2}{1-x}+\cdots=$$ $$\frac{1}{1-x}\cdot (1+ax+a^2x^2+\cdots)=\frac{1}{1-x}\cdot \frac{1}{1-ax}=\frac{1}{(1-x)(1-ax)}.$$ Method 2: Use sigma notations: $$\sum_{i=1}^{\infty} x^i \sum_{j=1}^i a^j=\sum_{i=1}^{\infty} x^i \cdot \frac{1-a^{i+1}}{1-a}=$$ $$\frac{1}{1-a}\sum_{i=0}^{\infty} x^i-\frac{a}{1-a}\sum_{i=1}^{\infty} (ax)^i=$$ $$\frac{1}{1-a}\cdot \frac{1}{1-x}-\frac{a}{1-a}\cdot \frac{1}{1-ax}=\frac{1}{(1-x)(1-ax)}.$$