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?




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