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 Sep 30 asked General question on relation between infinite series and complex numbers Sep 30 revised Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ added 515 characters in body Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ [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.] Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ @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]? Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ Srivatsan, +1. Thanks. Appreciate greatly. Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ @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. -cheers Sep 29 awarded Peer Pressure Sep 29 revised Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ added 241 characters in body; edited title Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ @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.) :) Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ @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. Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ @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. Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ @Fredrik: Can you provide a full epsilon-delta proof for this? This is where I'm stuck [for first part]. Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ @J.M.: I need a sort of epsilon-delta proof if I'm not mistaken. Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ [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. Sep 29 comment Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ 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. ... Sep 29 asked Show that the geometric series $a + ar +ar^2 + \cdots + ar^{n-1} + \cdots$ converges if and only if $|r| < 1$ Sep 17 comment Primer on complex analysis and Riemann surfaces for undergraduate physics / theoretical physics majors Ok. Thanks. - regards Sep 17 comment Primer on complex analysis and Riemann surfaces for undergraduate physics / theoretical physics majors Ref: math.stackexchange.com/questions/tagged/education ... Plus, the description field is empty. ... Sep 17 comment Primer on complex analysis and Riemann surfaces for undergraduate physics / theoretical physics majors I've looked through the posts with that tag. I found mostly questions similar to mine. ... I think the posts with that tag are different for the two sites. ... (Need clarification here.) Sep 17 revised Primer on complex analysis and Riemann surfaces for undergraduate physics / theoretical physics majors edited tags