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What's wrong with this proof? 1+2+4+... = -1

What's wrong with this proof?

Let \(S = 1 + 2 + 4 + 8 + ... \)

Then note that \(S = 1 + 2 + 4 + 8 + ... = 1 + 2(1 + 2 + 4 + 8 + ...) = 1 + 2S\)

And thus \(S = -1\).

Note by Zi Song Yeoh
4 years, 9 months ago

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Essentially it has to do with the fact that you're ignoring the last term, which is infinitely large and therefore very important. Let's consider this instead: \(S_n = 1+2+4+8 + \ldots + 2^{n+1} = 1 + 2(1+2+ \ldots + 2^{n}) = 1+ 2\cdot S_{n-1}\)

Take \(S_1 = 7\) and note that \(S_n = 1+ 2\cdot S_{n-1} > 2\cdot S_{n-1}\), so clearly \(S_n > 7 \cdot 2^{n-1}\) for \(n>1\). Hence your sum diverges (taking \(n\) large).

In your solution, you're taking \(S_n = S_{n-1}\), but we in fact know that \(S_n - S_{n-1} = 2^{n+1}\), so you're treating \(2^{n+1}\) as negligible for large \(n\). This is generally a bad idea.

Eli Ross - 4 years, 9 months ago

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The first issue is whether or not \(S\) makes sense in this context. Remember that the order of operations are functions \( +, -, \times, \div : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}\). Composition of functions is only defined for finite compositions, which means that we can only sum up terms finitely many times. Analysis is concerned with rigourizing what it means to be the sum of an infinite series. For example, we have all heard that for a Geometric Progression, we have a "sum to infinity" which has value \( \frac {a}{1-r} \), subject to the condition that \( |r| < 1\). In general, infinite sums can be tricky to understand. For example, what is the value of the infinite sum \( 1 -1 + 1 - 1 + 1 - 1 + 1 .... \)? Is it 0, or 1? Is it both? If so, is this a proof that \(0=1\)?

Let's assume that we dealt with the first issue, and have properly defined \(S\) to be \(\infty\), whatever that means.

The second issue, is whether we can extend the domain of our operations \(+, -, \times, \div\) to include \( \infty\). Sadly, this is not the case, especially when \( -, \div\) are involved. We cannot say that \( \infty - \infty = 0 \), or that \( \infty \div \infty = 1 \). We will need to learn how to extend this on a case by case basis. In fact, we will no longer have a (single-value) function, and \( \infty - \infty\) could actually be any value, from \( -\infty\) to \(\infty\)!

Calvin Lin Staff - 4 years, 9 months ago

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Nothing is wrong if you are looking in the 2-adic numbers :P

Lawrence Sun - 4 years, 9 months ago

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My guess is that S is not really a number in the normal sense; it is ∞, which doesn't really follow these kinds of rules. If ∞ + ∞ = ∞, can we say that ∞ = 0? No. I'm not sure how you would say this in formal mathematical terms though.

Tony Jiang - 4 years, 9 months ago

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infinite sum of a gp (geometric progression) is only valid when the common ratio is less than one and greater than minus one, and thus making the desired result. but when the common ratio is greater than one or less than minus one, the last term of an infinite gp is either infinity or minus infinity and thus the sum would amount to each of infinity or minus infinity and the method would not be valid...

Indrasis Roy - 4 years, 9 months ago

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sum to an infinite geometric series a/(1-r)is valid only when -1<r<1 . That is if the series is conversing to 0 ,then we can use a/(1-r). Given series is diverging in nature. So, it is wrong

Sanjiv Kumar - 4 years, 9 months ago

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Even when a sum converges in a special way, we might run into issues with how we consider \(S\). This "convergence", called "conditionally convergent", means that while \(\lim_{n\rightarrow\infty} \sum_{k=1}^{n} a_k\) indeed converges, \(\lim_{n\rightarrow\infty} \sum_{k=1}^{n} \left|a_k\right|\) does not. As one example, we often consider the sum \(\lim_{n\rightarrow\infty} \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} = \ln(2)\). But, if we rearranged the terms in the summation from \(\frac{{(-1)}^{k+1}}{k}\) for \(k=1,2,\ldots\) to \(\frac{1}{2j-1} - \frac{1}{4j-2} - \frac{1}{4j}\) for \(j=1,2,\ldots\), we would have the "same" terms in the summation, yet get a sum of \(\ln\left(\sqrt{2}\right)\). While it's a bit more complicated than what you've asked/we're talking about here, it's a pretty interesting phenomenon, and certainly related. You can read some more about it here.

To elaborate on the last point with a little less notation, here is why this dilemma is analogous to yours, even when this sum converges:

\(S = 1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots\)

Also, rearranging terms:

\(S = 1- \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \frac{1}{12} + \ldots\)

But since \(\frac{1}{2j-1} - \frac{1}{2(2j-1)} = \frac{1}{2(2j-1)}\), this is also:

\(S = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \ldots = \frac{1}{2} \cdot \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots\right)\).

So it appears that \(S = \frac{1}{2} S\), which seems not to make sense since we determined that \(S\) was \(\ln(2)\). While your solution makes the error in assuming \(S_n = S_{n-1}\) (see my post three below), here we have made the error in assuming that a rearranged sum \(S'\) must satisfy \(S' = S\). This is an example of why considering indices of infinite sums/order of summation - even for converging sums where those trailing terms are very small - is so critical.

Eli Ross - 4 years, 9 months ago

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I feel that the series being a geometric progression and common ratio being 2, which obviously is greater than one, sum is taken as infinite. Now infinite - infinite can be any real number or positive or negative infinite, while here it is zero.

Aaditya Gupta - 3 years, 6 months ago

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@Anas E. - It's exactly 1. The number \(0.\overline{9}\) is the number with infinitely many 9s, so you are essentially taking the limit. Further discussion/clarification can be found here.

Eli Ross - 4 years, 9 months ago

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hi

but it's "near" to 1, but it's not 1

like f(x)=\(\frac{1}{x}\)

for x= \(\infty\), f(x) is reaaally near to 0, but it's in NOT 0.

Anas Elidrissi - 4 years, 9 months ago

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@Anas E. - In fact, that's a valid use of this method. We indeed have \(0.\overline{9} = 1\). It's similar to Zi Song Y.'s, except that \(S_n - S_{n-1} = \frac{9}{10^n}\), so for large \(n\), that last term IS negligible, and taking the limit you arrive at the result you found.

Eli Ross - 4 years, 9 months ago

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From the geometric progression, sum(which is calculate by the sigma). meanwhile the unknown goes to the unlimited, the answer cannot be come out as such an specific number

Wonil Lee - 4 years, 9 months ago

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here's an other example

let x=0,9999999...

then 10x=9,99999...

therefor 9x=9

so x=1!!!

Anas Elidrissi - 4 years, 9 months ago

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But this is true

Shourya Pandey - 4 years, 7 months ago

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in fact we have \( 2^{0} + 2 +2² \) ... \( 2^{n}= 2^{n+1} -1\) and for each intenger a, \( a^{0} + a +a² \)... \(a^{n}\) = \( \frac{ a^{n+1} -1}{a-1} \)

Anas Elidrissi - 4 years, 9 months ago

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I was taught that a diverging series like this couldn't be written in terms of S = (...). Rather, as a limit, like \(\lim_{n \to \infty} \sum_{i=0}^{n}2^{i}\). Even \(\sum_{i=0}^{\infty}2^{i}\) was improper. \(\infty\) isn't really a number; it's like a "concept" that can only be approached. At least that's what I learned in school. =P

Tony Jiang - 4 years, 9 months ago

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Ya, this is what I thought.

There should be something to do with the \(\infty\)

But I'm not so sure...

Zi Song Yeoh - 4 years, 9 months ago

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