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In my econometric book, it says

$\sum\limits_{i=1}^n a^{i-1}=(1-a^n)/(1-a)$ because $\sum\limits_{i=1}^n a^{i-1}=(1-a^n)\sum\limits_{i=0}^\infty a^i$.

I understand the proof of $\sum\limits_{i=0}^\infty a^i=1/(1-a)$ and $\sum\limits_{i=1}^n a^{i-1}=(1-a^n)/(1-a)$ separately. But how do we know $\sum\limits_{i=1}^n a^{i-1}=(1-a^n)\sum\limits_{i=0}^\infty a^i$, assuming we don't know the representation of $\sum\limits_{i=0}^\infty a^i$ and $\sum\limits_{i=1}^n a^{i-1}$?

Q.L.
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  • I wonder about how the book proves that $\sum_{h=0}^\infty a^k=\frac1{1-a}$ for all $\lvert a\rvert<1$ without the formula for the partial sums. –  Sep 07 '18 at 00:56
  • @SaucyO'Path NO! That's not what I am asking. I understand each proof in the link you gave. The problem is how to link these two series before knowing these proofs. In the book, the logics is this link can lead to the representation of $\sum a^{i-1}$. – Q.L. Sep 07 '18 at 00:56
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    @SaucyO’Path no it is not a duplicate of that, the OP here is asking for something related to that but different – user Sep 07 '18 at 00:57
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    @SaucyO'Path Is that really a duplicate? The OP said he knows those proofs in the original post. I think people here just love closing questions. – Ryan Sep 07 '18 at 00:57
  • @Ryan Based on what I can see, the argument presented in the OP is circular and it deals with just a very small number of cases and it doesn't mention any of these facts. Now, here comes a phylosophical dilemma: is the logic of flawed versions of a concise proof worth an analysis, or is it better to just state a correct proof and show how easy it is? People being people, they like giving different answer. –  Sep 07 '18 at 01:12
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    @SaucyO'Path The OP asked for a clarification on a statement about these series framed in a certain way. You linked to the proofs the OP stated he knows. I don't understand how linking to proofs that the OP stated he/she knows helps at all, and initiating to close the question like that is actually the opposite of helping! – Ryan Sep 07 '18 at 01:31

1 Answers1

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It should be true only for $|a|<1$ since

$$\sum\limits_{i=1}^n a^{i-1}=\frac{1-a^n}{1-a} \implies \sum\limits_{i=0}^\infty a^i=\frac1{1-a}$$

and therefore

$$\sum\limits_{i=1}^n a^{i-1}=(1-a^n)\sum\limits_{i=0}^\infty a^i=\frac{1-a^n}{1-a}$$

otherwise the equality doesn’t hold since the series doesn’t converge.

user
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