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Show, rigorously, that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if |r| < 1.

Also, show that if |r| < 1, the sum is given by $S = \frac{a}{1-r}$.


Note(s):

Question 1, First part: if |r| < 1, show (rigorously) that the geometric series above converges.

Update: I'm reposting this question as the accepted answer on the possible duplicate question doesn't provide a rigorous [$\epsilon$-$\delta$] derivation of an important step, namely, $\lim_{N\to\infty}r^{N+1} = 0$ if $|r|\lt 1$. [Translated to the current question's context, $\lim_{n\to\infty}ar^n = 0$ if $|r|\lt 1$].

Update 2: The problem of the above issue has been addressed.


Question 1, Second part: if the geometric series above converges, then show (rigorously) that |r| < 1.

-- Any hints would be appreciated. --

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    What have you tried? What do you think happens to the terms of the series if $|r| \ge 1$?2011-09-29
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    You should add: "or $a=0$".2011-09-29
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    Another one: if the series has a sum $S$, then $S=a+rS$ (why?).2011-09-29
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    What is the sum of the first $n$ terms? (You should know the formula for the sum of a finite geometric series.) What happens to that sum as $n\to\infty$? How does the answer to that question depend on $a$?2011-09-29
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    Basically, first I derive the formula for the finite series as $S_n = a \frac{(r^n -1)}{r-1}$; then, I claim that, if |r| < 1, convergence of the series is equivalent to claiming that: lim$_{n \rightarrow \infty}$ |$S_{n+1} - S_n$| = lim$_{n \rightarrow \infty}$ |$ar^n$| $\rightarrow$ 0. But I'm wondering how to prove this. ...2011-09-29
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    You can factor out the $a$ from the last limit. Now, would you happen to remember how $a^x$ behaves for both increasing and decreasing $x$?2011-09-29
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    [Derivation for $S_n = a \frac{(r^n-1)}{r-1}$ was by simple algebraic manipulation. I don't need any assistance with that.] ... The specific issue I have with first part of first proof is, how to prove above claim.2011-09-29
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    Clearly if $|r| < 1$, then $r^n \to 0$. This is because, for any $0 < \epsilon < 1$, you can solve $r^n=\epsilon$ using the logarithm.2011-09-29
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    @J.M.: I need a sort of epsilon-delta proof if I'm not mistaken.2011-09-29
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    Once you have the partial sums, you don't need to look at the difference and show when it goes to zero. You can just directly look at its limit behavior, because that's how $S$ is defined.2011-09-29
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    @Fredrik: Can you provide a full epsilon-delta proof for this? This is where I'm stuck [for first part].2011-09-29
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    @UGPhysics: Your question seems to have morphed into a different question. Your original question is answered in the post I suggested as a duplicate above. If you have a new question specifically about a certain detail of the answers there, I think you should rewrite the question or open a new one; in its current form, the question is a duplicate of the other one.2011-09-29
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    @wnoise: I need to demonstrate that the series converges; if it does, the sum has meaning. I can't use the concept of sum if I cannot first demonstrate that the series converges.2011-09-29
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    @UGPhysics: That last comment is incorrect; wnoise's comment was correct. Partial sums exist independent of the convergence of the corresponding series, and their convergence can be analyzed. The value of a series is nothing but the limit of the partial sums.2011-09-29
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    Your comment #7 shows a lack of understanding. It's not enough to show that $S_{n+1}-S_n$ tends to $0 -$ take for instance $S_n=\sum_{r=1}^n 1/r$, which doesn't converge.2011-09-29
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    @UGPhysics Here goes: You want to show that $r^n \to 0$ as $n \to \infty$ provided $|r| < 1$. So let $\epsilon > 0$ be given. Then $r^n \leq \epsilon$ is equivalent to $n \geq \ln \epsilon/ \ln r$. Such an $n$ exists.2011-09-29
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    @wnoise: by using the derived formula (partial sum formula) I then claim that if |r| < 1, the expression tends to a limiting values (i.e. 'sum' S) as successive differences tend to zero as n tends to infinity. I need to prove that successive differences (|$S_{n+1} - S_n$|) tend to zero (for n $\rightarrow \infty$). .. I'm currently trying to find the exact epsilon-delta argument to demonstrate this.2011-09-29
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    @Fredrik: I've never done a formal $\epsilon - \delta$ proof before. Can you provide a detailed version for a complete newbie to fully understand. ... (Would appreciate greatly.) :)2011-09-29
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    @UGPhyiscs: If by "reposting" you mean posting this same question again, please don't do that; it was rightly closed as a duplicate. You are of course free to post a new question asking specifically about an $\epsilon$-$\delta$ proof of the convergence of $r^n$.2011-09-29
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    You don't prove that a sequence converges with epsilon and deltas; proofs of convergence of *series* don't use deltas.2011-09-29
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    To expand on @Arturo's comment, to prove the convergence of sequences or series, you use epsilon and $N$ (or $N_0$ or $N_\epsilon$, different people use various notations), but not $\delta$. That is, a sequence $s_n$ converges to $L$ iff given $\epsilon > 0$, there exists a natural number $N$ such that we have $|s_n - L| \leq \epsilon$ for all $n \geq N$.2011-09-29
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    @Arturo: Can you provide your perspective on how best to prove the above [in question body]? I think my general approach is in the right direction, but maybe you can suggest a better method / methods. -cheers2011-09-29
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    Srivatsan, +1. Thanks. Appreciate greatly.2011-09-29
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    @UGPhysics: If, as you claim, the issue here is a proof that $\lim_{n\to\infty}r^n = 0$ when $|r|\lt 1$, 1 when $r=1$, and does not exist in all other cases, then I do not see any approach that you have to do that. As such, I cannot comment if it is "in the right direction", since it does not seem to exist at all. In fact, your continued requests about "$\epsilon-\delta$ proof" suggests you haven't even thought about what a "formal proof of the limit" means, let alone how to approach obtaining such a proof.2011-09-29
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    @UGPhysics This is my recommendation. First, your question naturally reduces to showing that $r^n$ converges to $0$ iff $|r| < 1$ (perhaps using epsilon-$N_0$ arguments). I guess it is ok if you post *this* as a separate question. You should be able to fill in the remaining steps either by yourself or with the help of answers from the other question.2011-09-29
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    @Arturo: The issue of the epsilon-delta proof comes from the following: [comment # 5] "first I derive the formula for the finite series as $S_n = a\frac{(r^n−1)}{r−1}$; then, I claim that, if |r| < 1, convergence of the series is equivalent to claiming that: $\lim_{n \rightarrow \infty} |S_{n+1}−S_n| = \lim_{n \rightarrow \infty} |ar^n| = 0$. But I'm wondering how to prove this." ... I'm wondering if this is a valid approach to the initial problem [the first part of]?2011-09-29
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    [Plus, as per FredrikMeyer's comment ["You want to show that $r^n \rightarrow 0$ as $n \rightarrow \infty$ provided |r|<1. So let ϵ > 0 be given. Then $r^n \lt ϵ$ is equivalent to n ≥ ln ϵ/ ln r. Such an n exists."] I'm not sure why you would claim a "formal" $\epsilon$-$\delta$ proof of the statement would not be applicable.]2011-09-29
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    @UGPhysics I already explained Arturo's comment that epsilon-*delta* proofs do not apply here (only epsilon-$N_0$ proofs). Your subsequent comments on epsilon-delta make me doubt that you understand the idea of doing epsilon-delta or epsilon-$N_0$ proofs.2011-09-30
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    @UGPhysics: In any case, I have added a proof of this limit in [the original answer](http://math.stackexchange.com/questions/29023/value-of-sum-xn/29035#29035). It uses the infimum property, the Archimdean property, and that a subsequence of a converging sequence must converge to the same limit as the original sequence. The latter is very easy; the previous two would require a formal construction of the real numbers in order to provide a formal proof.2011-09-30

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