Prove that $\left({1 \over 1-x}\right)^2 = \sum_{n=0}^\infty (n+1)x^n$

Question

Prove that if $x\in [0, 1)$, then $\displaystyle \left({1 \over 1-x}\right)^2 = \sum_{n=0}^\infty (n+1)x^n$

I have a theorem which claims that

THEOREM: If for some $x\in\mathbb R$ the power series $\sum_{n=0}^\infty a_nx^n$ and $\sum_{n=0}^\infty b_nx^n$ are absolutely convergent, then $$\left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{n=0}^\infty b_nx^n\right) = \left(\sum_{n=0}^\infty c_nx^n\right),$$ where $c_n = \sum_{k=0}^na_kb_{n-k}.$

I attempted the proof and the heart of the proof is extremely straightforward: $$\begin{align}\left({1 \over 1-x}\right)^2 & = \left({1 \over 1-x}\right)\left({1 \over 1-x}\right)\\ & = \left(\sum_{n=0}^\infty1x^n\right)\left(\sum_{n=0}^\infty1x^n\right) \tag{Geometric Series}\\ &= \sum_{n=0}^\infty\left(\left[\sum_{k=0}^n(1)(1)\right]x^n\right) \tag{Theorem}\\ &=\sum_{n=0}^\infty\left[(n+1)x^n\right].\end{align}$$

However, is it necessary for the proof to show that $\sum_{n=0}^\infty x^n$ is absolutely convergent? I'm not entirely sure how to go about doing that. If I had to hazard a guess, I would say that since $x \in [0, 1)$, then $|x^n| = |x|^n = x^n$ for all $n\in\mathbb N$, so $\sum_{n=0}^\infty |x^n| = \sum_{n=0}^\infty x^n$ is convergent since it is a geometric series which implies the series is absolutely convergent. But is it valid to say two infinite series are "equal" like this?

Answer 1

Although you have the multiplication theorem at hand, but you can easily prove the problem by differentiation. That is, try to differentiate both sides of the following identity $$\frac1{1-x}=\sum_{n=0}^\infty{x^n}$$ and you'll get this $$\left(\frac1{1-x}\right)^2=\sum_{n=0}^\infty{nx^{n-1}}$$ Since the first term of the summation is zero, you can change the index to get this $$\left(\frac1{1-x}\right)^2=\sum_{n=0}^\infty{(n+1)x^{n}}$$

Answer 2

Your proof is correct. The geometric series converges absolutey for $|x|<1$.

So, what is your problem ?

Answer 3

If you are allowed to start from the right, the problem is simple since $$\sum_{n=0}^\infty (n+1)x^n=\sum_{n=0}^\infty nx^n+\sum_{n=0}^\infty x^n=x\sum_{n=0}^\infty nx^{n-1}+\sum_{n=0}^\infty x^n=x \left(\sum_{n=0}^\infty x^n \right)'+\sum_{n=0}^\infty x^n$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \pars{1 \over 1 - x}^{2} & = \pars{1 - x}^{-2} = \sum_{n = 0}^{\infty}{-2 \choose n}\pars{-x}^{n} = \sum_{n = 0}^{\infty}\braces{{-\bracks{-2} + n - 1 \choose n}\pars{-1}^{n}}\pars{-x}^{n} \\[5mm] & = \sum_{n = 0}^{\infty}{n + 1 \choose n}x^{n} = \bbx{\ds{\sum_{n = 0}^{\infty}\pars{n + 1}x^{n}}} \end{align}