How to prove the closed form $\left(\frac {1}{1-x}\right)^2 = \sum_{n=0}^{\infty}(n+1)x^n$

Question

Show for $x \in \mathbb{R}$ with $|x|< 1 $

$$\left(\frac {1}{1-x}\right)^2 = \sum\limits_{n=0}^{\infty}(n+1)x^n$$

My idea is, to use the Cauchy product, but i didn't gone very far..

Answer 1

HINT: Start with the geometric series $$\frac1{1-x}=\sum_{n\ge 0}x^n$$ and differentiate.

Answer 2

If you don't like differentiation, letting $S$ denote the quantity you are interested in, you can compute $$ S - xS = \sum_{n=0}^\infty (n+1)x^n - \sum_{m=0}^\infty (m+1) x^{m+1} = \sum_{n=0}^\infty (n+1)x^n - \sum_{n=1}^\infty n x^{n} $$ where in the second summation I made the substitution $n = m+1$.

Answer 3

From geometric series, we have $$\sum_{n=0}^{\infty}x^n = \dfrac1{1-x}$$ for $0 < \vert x \vert < 1$. Differentiating both sides, we get that $$\sum_{n=0}^{\infty}nx^{n-1} = \dfrac1{(1-x)^2}$$

Answer 4

Viewing $\frac1{1-x}=\sum_{n=0}^{\infty}x^n$ as a formal power series, it is (also) easy to compute its square: the coefficient of $x^n$ in $(\frac1{1-x})^2$ is the number of solutions in non-negative integers $i,j$ of $i+j=n$ (since each term $x^i$ multiplied by a terms $x^j$ gives a term $x^{i+j}$), which number of solutions is $n+1$. Since the growth of the coefficient $n+1$ is less than $\alpha^n$ for any $\alpha>1$, the series $\sum_{n=0}^{\infty}(n+1)x^n$ is absolutely convergent for any $x\in\Bbb C$ with $|x|<1$ .

Noting that the square of $\sum_{n=0}^{\infty}x^n$ is the same as its derivative, you may in passing observe to have found a solution of the differential equation $y'=y^2$ (with $y(0)=1$).