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When is the formula $$S_{\infty} = \dfrac{a}{1-r}$$ valid?

Does |$r| <1$?

Phaptitude
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    The equality holds for $-1<r<1$ and is undefined otherwise. – Git Gud Oct 27 '13 at 22:51
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    @GitGud You should post it as an answer, perhaps with a little bit of motivation too, if you want, so I can thank you properly – Phaptitude Oct 27 '13 at 22:52
  • I'm off soon. This is explained everywhere, though. For instance here. – Git Gud Oct 27 '13 at 22:55
  • @GitGud Yeah, I usually mess around with this stuff, forgetting that is probably answered elsewhere. I'll try to keep it to a minimum in the future. – Phaptitude Oct 27 '13 at 22:57
  • Yes, it's valid if and only if $|r| < 1$. This includes the case when $r$ is complex and $|r|$ denotes modulus as opposed to absolute value. If $a = 0$, then for any $r$, all the terms in the series are zero and the series converges to zero, but I don't think the series is considered a geometric series. – Stefan Smith Oct 27 '13 at 23:52

2 Answers2

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Yes, indeed, you are correct in your suspicion: So long as $\; −1<r<1,\;$ the formula $$ S_\infty = \sum_{k=0}^\infty ar^k = \frac{a}{1-r}$$

holds for a geometric series.

amWhy
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Since I found this very useful when learning geometric series, I'll show you this to enhance your understanding of the formula:

Given a finite geometric sequence, let's say exempli gratia that it has 5 terms. So the terms are $a, ar, ar^2, ar^3, ar^4$. We want to find the sum of this sequence:

$x = a + ar + ar^2 + ar^3 + ar^4$

Multiply through by r:

$xr = ar + ar^2 + ar^3 + ar^4 + ar^5$

Subtract (2) from (1) to get $x(1 - r) = a - ar^5 \rightarrow x = \frac {a(1 - r^5)}{1-r}$

So that's why the formula is why it is. Thus, when you have an infinite geometric series, it only holds when $\lim_{n \rightarrow \infty} r^n = 0$, which occurs when $|r| < 1$

MT_
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