Prove that (11−x)2=∑∞n=0(n+1)xn\left({1 \over 1-x}\right)^2 = \sum_{n=0}^\infty (n+1)x^n

Prove that if x∈[0,1)x\in [0, 1), then (11−x)2=∞∑n=0(n+1)xn\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∈Rx\in\mathbb R the power series ∑∞n=0anxn\sum_{n=0}^\infty a_nx^n and ∑∞n=0bnxn\sum_{n=0}^\infty b_nx^n are absolutely convergent, then (∞∑n=0anxn)(∞∑n=0bnxn)=(∞∑n=0cnxn),\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 cn=∑nk=0akbn−k.c_n = \sum_{k=0}^na_kb_{n-k}.

I attempted the proof and the heart of the proof is extremely straightforward: (11−x)2=(11−x)(11−x)=(∞∑n=01xn)(∞∑n=01xn)=∞∑n=0([n∑k=0(1)(1)]xn)=∞∑n=0[(n+1)xn].\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 ∑∞n=0xn\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∈[0,1)x \in [0, 1), then |xn|=|x|n=xn|x^n| = |x|^n = x^n for all n∈Nn\in\mathbb N, so ∑∞n=0|xn|=∑∞n=0xn\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?

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You don’t need that. Just multiply both sides of 1(1−x)2?=∑n≥0(n+1)xn\frac{1}{(1-x)^2}\stackrel{?}{=}\sum_{n\geq 0}(n+1)x^n by (1−x)2(1-x)^2 and notice a massive cancellation in the resulting RHS.
– Jack D’Aurizio
2 days ago

  

 

But, of course, convolution is a good way through. The Taylor series of \frac{1}{1-x}\frac{1}{1-x} is absolutely convergent as soon as |x|<1|x|<1, so it is its square. – Jack D'Aurizio 2 days ago ================= 3 Answers 3 ================= 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}\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}}\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}}\left(\frac1{1-x}\right)^2=\sum_{n=0}^\infty{(n+1)x^{n}}      This is a very good answer and it is how I first approached the problem. However, it is my fault that I probably should have specified that no such thing as differentiation has been mentioned in my textbook just yet. Therefore, we must use other tools to compute this (hence why I used a corollary to the Cauchy product). – pyrazolam 2 days ago 1   I see. In order to prove the absolute convergence of the geometric series, you can start by deriving the finite sum formula, i.e., since x\in[0,1)x\in[0,1) s_N=\sum_{n=0}^N{|x^n|}=\sum_{n=0}^N{x^n}=\frac{1-x^N}{1-x}s_N=\sum_{n=0}^N{|x^n|}=\sum_{n=0}^N{x^n}=\frac{1-x^N}{1-x} and then show its limit exists as NN approaches infinity. – Babak 2 days ago      although the answer itself is not as satisfactory as I'd like, I will choose this answer as correct for the comment listed above. 🙂 – pyrazolam 2 days ago      Thank you. And please consider this correction s_N=\frac{1-x^{N+1}}{1-x}s_N=\frac{1-x^{N+1}}{1-x}. – Babak 2 days ago 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\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 Your proof is correct. The geometric series converges absolutey for |x|<1|x|<1. So, what is your problem ?      The issue I'm battling is listed at the very bottom of the post--is it valid to say that \sum_{n=0}^\infty |x^n| = \sum_{n=0}^\infty x^n\sum_{n=0}^\infty |x^n| = \sum_{n=0}^\infty x^n since x\in[0,1)x\in[0,1), then |x^n| = x^n|x^n| = x^n for all n\in\mathbb Nn\in\mathbb N? It feels like sound logic, but my concern was that it felt like it's a very large jump in conclusions that \sum_{n=0}^\infty |x^n|\sum_{n=0}^\infty |x^n| converges in the first place--so would it even make sense to say it equals a convergent series? That was my issue. – pyrazolam 2 days ago      Yes you can! It's the same series exactly under a different name. – Stoke 2 days ago