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The geometric series $\sum_{k=0}^{\infty}2^{-k}$ converges to 2. Use Theorem 6.3.6 to show that the series $$\sum_{k=0}^{\infty}(k+1)2^{-k}$$ converges to 4.

Theorem 6.3.6 states: Let $\sum_{k=0}^{\infty}a_k$ and $\sum_{j=0}^{\infty}b_j$ be two absolutely convergent series. Then $$(\sum_{k=0}^{\infty}a_k)(\sum_{j=0}^{\infty}b_j)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_kb_{n-k}$$ where the series on the right also converges absolutely. From reading other questions, it looks like this is called the Cauchy product.

It seems like a good first step to assume $2^{-k}$ will be playing the role of $a_k$ here. Then I am looking for something to play the role of $b_j$ such that $\sum_{j=0}^{\infty}b_j$ converges to 2 and such that $k+1$ emerges from the $b_{n-k}$ terms.

I'm thrown off by the inner sum. It seems like something along the lines of $$b_j=\frac{(b+1)!}{b!}$$ could lead to the $k+1$ emerging, but this would make $\sum b_j$ a divergent series.

Is my choice of $a_k$ reasonable? Is there a different way I should be thinking about $b_j$?

This is not a duplicate of How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$? because this problem requires use of a theorem that is not used in the linked question.

Jacob
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  • See https://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-inftyn1xn – Travis Willse Nov 26 '19 at 02:56
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    just differentiate the geometric series and plug in a half – operatorerror Nov 26 '19 at 02:56
  • It's $ 2\sum\limits_{k=1}^\infty k(\frac12)^k$ – J. W. Tanner Nov 26 '19 at 02:57
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    Set $a_k = 2^{-k}$ and $b_k = 2^{-k}$. LHS: $4$, RHS: $\sum_{n=0}^{\infty}\sum_{k=0}^n 12^{-n}= \sum_{n=0}^{\infty}(n+1)2^{-k}$. Hence the answer is 4 – fGDu94 Nov 26 '19 at 03:08
  • This theorem doesn't seem appropriate because $\sum_{k=0}^\infty (k+1)2^{-k}$ can't really be written in the form $\left( \sum_{k=0}^\infty a_k\right)\left(\sum_{k=0}^\infty b_k\right)$. I would just use differentiation of power series and the knowledge of the limit of the geometric series here. – Math1000 Nov 26 '19 at 03:09
  • @qbert Unfortunately we haven't developed the tools to differentiate series yet in the course. I need to find a justification that non-trivially uses the statement in the theorem (even if simpler methods are available). – Jacob Nov 26 '19 at 03:12

2 Answers2

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Set $a_k = 2^{-k}$ and $b_k = 2^{-k}$.

LHS: $4$,

RHS: $\sum_{n=0}^{\infty}\sum_{k=0}^n 1*2^{-n}= \sum_{n=0}^{\infty}(n+1)*2^{-n}$. Hence the answer is 4

fGDu94
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  • Thanks! Is there a thought process that leads to seeing that this is the answer, or is this more of a 'flash of insight' solution? I understand why this works, but I'm not sure how I would go about seeing it if I didn't already know to try it. – Jacob Nov 26 '19 at 03:21
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    well basically it says use the theorem. You can't have $a_k$ or $b_k$ just be $(k+1)$ on its own as this does not sum up. Also, you can't have $a_k = (k+1)2^{-k}$ because you get square terms on the right hand side (oh no!). So I realised that this sum from 0 to n on the right hand side was just an $n+1$ in disguise – fGDu94 Nov 26 '19 at 03:23
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I think the intention here is to write

$$(n+1)2^{-n} = \sum_{k=0}^n\left(\frac 1{2^k}\cdot \frac 1{2^{n-k}} \right)$$

Then apply the given theorem about the Cauchy product.

trancelocation
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