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For any geometric series with |$r$| < 1 , I know that

$$\sum_{k=1}^{∞} ar^{k-1} =\frac{a}{1-r}$$

But if |$r$| > 1 and you try to use the formula, you'll get a weird answer. For instance:

$$4+8+16+32+64+128+... =\sum_{k=1}^{∞} (4)2^{k-1}= \frac{4}{1-2} = -4$$

That answer obviously doesn't make sense; the series diverges. So what does -4 mean? Where did it come from, and how is it related to the series? It must be significant somehow.

tethernova
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    Just a quick comment: The value $-4$ may somewhat be related to analytic continuation of the complex function $1/(1-z)$, just as a famous "formula" $1+1/2+1/3+\cdots=-1/12$, but I'm not sure at all. Someone (or I?) will discuss about it in answer when he/she gets convinced. – Katie Imach Dec 19 '15 at 04:50
  • Sorry for typos...I mean $1+2+3+\cdots=-1/12$. This is a consequence of analytic continuation of a Riemann zeta function. – Katie Imach Dec 19 '15 at 05:01
  • Maybe you want to read this? http://math.stackexchange.com/questions/39802/why-does-123-cdots-frac112 – MonkeyKing Dec 19 '15 at 05:07
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    Your summation can also be derived with some (precarious) algebra. What is $x$? $$\begin{align}x&=4+8+16+32+64+\cdots \ (1/4)x&=1+2+4+8+16+\cdots \ (1/4)x&=1+2+x \end{align}$$ – Corellian Dec 19 '15 at 05:12

4 Answers4

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For a truncated series, we have

$$\sum_{k=1}^Kar^{k-1}=a\frac{1-r^K}{1-r} \tag 1$$

For $a=4$ and $r=2$, use of $(1)$ reveals

$$\sum_{k=1}^K4\left(2\right)^{k-1}=4\frac{1-2^K}{1-2}=2^{K+2}-4$$

The $-4$ appears as the "convergent component" of the divergent series (as $K\to \infty$).

Mark Viola
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  • This is a answer that tells everything to need for tethernova to think (400+?) – Moti Dec 19 '15 at 04:20
  • @Moti I have no idea what you mean. Help me understand. And why did you single out this answer when others posted similar ones?? – Mark Viola Dec 19 '15 at 04:23
  • Because it clearly explains your false interpretation of the sum equation that you used which is an estimation if r is small. IT IS NOT INTENDED TO BE USED WITH LARGE r - as result you get an unexplained erroneous answer. – Moti Dec 19 '15 at 04:26
  • @Moti Tethernova knows that - they want to know if it has any meaning. – Deusovi Dec 19 '15 at 04:33
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    @Moti Why the offensive tone? Please refrain. Or do we need a moderator here? And I have not posted a "false interpretation of the sum equation." And it is valid for all $K$ and for all $r$ - even complex valued $r$. This is merely a truncated sum and clearly shows the source of the $-4$ as the "convergent" component of the divergent series. Do you understand now? – Mark Viola Dec 19 '15 at 05:03
  • I was not referring to you, as you may see from my first comment - so what exactly you talk about? @Deusovi from the way he presented the question it is not clear that he asks about the meaning. A great discussion of related issues may be found in the zeta function... – Moti Dec 19 '15 at 05:10
  • Why was this down voted? – Mark Viola Sep 13 '18 at 14:19
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Let's look at the formula for a finite geometric series first. If $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms, then your sum is equal to $a\frac{1-r^n}{1-r}$. Now, if $|r|<1$, we can see that as $n$ goes to infinity the $r^n$ term disappears, leaving the familiar formula of $a\frac{1}{1-r}$. So, the infinite geometric series sum formula makes the assumption that $|r|<1$.

With the finite geometric series sum formula, we can rewrite as follows:

$$a\frac{1-r^n}{1-r}$$

$$=a\frac{r^n-1}{r-1}$$

$$=\frac{a}{r-1}r^n - \frac{a}{r-1}$$

Remember, an infinite sum is just the limit as $n\to\infty$ of a finite sum. Now, we know that the infinite sum assumes that $r^n$ goes to 0. If we assume that the first term goes to 0 even though it doesn't, we end up with the second term of that last line: $$\frac{-a}{r-1}$$

If you look at the terms individually, you can see that as $n\to\infty$ the first term goes to infinity. In a sense, you can think of that number you get as "ignoring infinity's contribution" to the sum (though this idea is very informal). It's what the sum would have been if the first term disappeared like it does when $|r|\lt1$.

Deusovi
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-6

This is a way of assigning meaning to an otherwise divergent sum.

Look up "divergent series."

The subject is large.

marty cohen
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-6

This is an estimation! If you will look how this estimation is derived you will see immediately that the sum goes to infinity because of $r^n$ in the sum

Moti
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