Expansion of this expression

Question

Let $x$ be a real number in $\left[0,\frac{1}{2}\right].$ It is well known that $$\frac{1}{1-x}=\sum_{n=0}^{+\infty} x^n.$$ What is the expansion or the series of the expression $(\frac{1}{1-x})^2$? Many thanks.

Answer 1

Method 1 : Binomial Theorem for any index

$$(1-x)^{-n}= 1+nx+ \frac{n(n+1)}{2!}x^2+ \frac{n(n+1)(n+2)}{3!}x^3 \dots $$

Putting $n=1$ , we get :

$$\frac{1}{1-x} = 1+x+x^2+x^3+ \dots $$

Putting $n=2$ , we get :

$$\frac{1}{(1-x)^2} = 1+2x+3x^2+4x^3+ \dots $$

Method 2 : Derivatives

Let: $$f(x)=\frac{1}{1-x} = 1+x+x^2+x^3+ \dots $$

Take derivative w.r.t. $x$ both the sides; $$f'(x)=\frac{1}{(1-x)^2} = 1+2x+3x^2+4x^3+ \dots $$

$$ \implies \frac{1}{(1-x)^2} = 1+2x+3x^2+4x^3+ \dots $$

NOTE : These expansions are valid not only for $ x \in \left[0,\dfrac{1}{2}\right]$ but $\forall ~x \in (-1,1)$

Answer 2

If you consider $$1+2x+3x^2\ldots \infty,$$ then it is $\dfrac{1}{{(1-x)^2}}$.

Proof: Let $S=1+2x+3x^2\ldots \infty \Rightarrow Sx=x+2x^2+3x^3\ldots\infty$ Now, $S-Sx=1+x+x^2\ldots\infty=\displaystyle\sum_{n=0}^{\infty}x^n=\dfrac{1}{1-x}$

$$S(1-x)=\dfrac{1}{1-x} \Rightarrow S=\dfrac{1}{{(1-x)^2}}.$$

$\therefore$ The series of the expansion is $S=\displaystyle\sum_{n=0}^{\infty}(n+1)x=\dfrac{1}{{(1-x)^2}}$, where $|x|<1$. This is just the Generating Functions.

Answer 3

By the Cauchy product,

$$\left(\sum_{n=0}^{\infty} x^n\right)^2=\left(\sum_{n=0}^{\infty} x^n\right)\left(\sum_{m=0}^{\infty} x^m\right)=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}x^nx^m=\sum_{k=0}^{\infty}\sum_{m=0}^k x^k=\sum_{k=0}^{\infty}(k+1)x^k.$$

Answer 4

Given that $$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$ for all $x\in(-1,1)$, we may differentiate both sides to obtain $$\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty}nx^{n-1} = \sum_{n=0}^{\infty}(n+1)x^n$$ for all $x\in(-1,1)$.

Here we have used the fact that a power series is differentiable inside it's interval of convergence, and the derivative can be obtained by differentiating term-by-term, just as you would for a polynomial.

Answer 5

Or do by brute force multiplication and gathering terms: $$\frac{1}{(1-x)^2}=$$ $$ (1+x+x^2+x^3+x^4+\ldots)(1+x+x^2+x^3+x^4+\ldots)\\ =(1+x+x^2+x^3+x^4+\ldots)\\ +x+x^2+x^3+x^4+\ldots\\ \quad +x^2+x^3+x^4+\ldots\\ \quad \quad +x^3+x^4+\ldots\\ =1+2x+3x^2+4x^3+\ldots $$