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What does \(1 + 2 + 4 + 8 + 16 + 32 + ...\) equal?
You could rewrite the equation as (2−1)(1+2+4+8+16+32+...)(2 - 1)(1 + 2 + 4 + 8 + 16 + 32 + ...)(2−1)(1+2+4+8+16+32+...)
This is then written as (2+4+8+16+32+...)−1−2−4−8−16−32−...(2 + 4 + 8 + 16 + 32 + ...) - 1 - 2 - 4 - 8 - 16 - 32 - ...(2+4+8+16+32+...)−1−2−4−8−16−32−....
Which results in 1+2+4+8+16+32+...=−11 + 2 + 4 + 8 + 16 + 32 + ...= -11+2+4+8+16+32+...=−1!!!.
Note by Sharky Kesa 7 years, 2 months ago
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This is an infinite geometric series which has ratio r=2>1r=2>1r=2>1. So this doesn't converge, and as the result we get an non-sense answer.
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I'm sorry but you can't just use the distributive laws when dealing with infinitely many terms. A simple proof that this is wrong is that a sum diverges when its partial sums diverges. Clearly the partial sums are 2n−12^n-12n−1 which approaches to infinity as nnn does, therefore the sum does not exist and is undefined.
Or instead you are talking about something much more complicated like in a different number system, like 2-adic numbers. :P
Damn, we communicated before Schol of Excellence???
I do think this answer is a bit weird, but the method seems correct, since it is an infinite series! and does not end.
Just an illusion.....,
Confine up to 5 or 6 terms and try to evaluate..
I had a similar derivation which I am not so sure about. My first note on Summations
these laws are not valid when we are dealing with infinite sums.
ahh I like these fallacious proofs - they really test your observation. I think the reason why you got -1 was because the factorization only works if the expression is finite.
It has to do with the fact that you're ignoring the last term, which is infinitely large and therefore very important
There is no largest term.
I do not see where −1-1−1 came from.
EDIT: For a proof of where this summation went wrong, check my comment down here.
Remove the brackets and you get 2+4+8+16+32+...-1-2-4-8-16-32. The +2 and -2 cancel each other out, so does +4 and -4, +8 and -8, etc. until all that's left is -1
This is obviously false, we still have a term in the first sequence missing.
If the RHS ends, let's say, in 2n2^n2n, the LHS will end in 2n+12^{n+1}2n+1. This leads us that the Σ=2n+1−1 \Sigma = 2^{n+1} - 1Σ=2n+1−1, which is true.
But we cannot evaluate a limit for n→∞n \rightarrow \inftyn→∞, because Σ\SigmaΣ would tend also to infinity.
@Guilherme Dela Corte – Since it was (2−1)(1+2+4+8+16+32+...)(2 - 1)(1 + 2 + 4 + 8 + 16 + 32 + ...)(2−1)(1+2+4+8+16+32+...), you just simplify it. 2(1+2+4+8+...)−1(1+2+4+8+16+...)2(1 + 2 + 4 + 8 + ...) -1(1 + 2 + 4 + 8 + 16 +...)2(1+2+4+8+...)−1(1+2+4+8+16+...) so it is 2+4+8+16+32...−1−2−4−8−16...2 + 4 + 8 + 16 + 32 ...- 1 - 2 - 4 - 8 - 16...2+4+8+16+32...−1−2−4−8−16.... 222 and −2-2−2, 444 and −4-4−4, etc. all cancel each other out until −1-1−1 remains
@Guilherme Dela Corte – However, the proof for 1+2+3+4+5+6+…∞=−1121+2+3+4+5+6+\ldots\infty=\frac{-1}{12}1+2+3+4+5+6+…∞=12−1, requires the assumption that the R.H.S. does not end. The point of infinite sequences is that they do not end. Even the proof for 9.99999…=109.99999\ldots=109.99999…=10 requires the 999s to never end.
Nice logic!!
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Top NewestThis is an infinite geometric series which has ratio r=2>1. So this doesn't converge, and as the result we get an non-sense answer.
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I'm sorry but you can't just use the distributive laws when dealing with infinitely many terms. A simple proof that this is wrong is that a sum diverges when its partial sums diverges. Clearly the partial sums are 2n−1 which approaches to infinity as n does, therefore the sum does not exist and is undefined.
Or instead you are talking about something much more complicated like in a different number system, like 2-adic numbers. :P
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Damn, we communicated before Schol of Excellence???
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I do think this answer is a bit weird, but the method seems correct, since it is an infinite series! and does not end.
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Just an illusion.....,
Confine up to 5 or 6 terms and try to evaluate..
InitiallyTrolled me too..,
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I had a similar derivation which I am not so sure about. My first note on Summations
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these laws are not valid when we are dealing with infinite sums.
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ahh I like these fallacious proofs - they really test your observation. I think the reason why you got -1 was because the factorization only works if the expression is finite.
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It has to do with the fact that you're ignoring the last term, which is infinitely large and therefore very important
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There is no largest term.
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I do not see where −1 came from.
EDIT: For a proof of where this summation went wrong, check my comment down here.
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Remove the brackets and you get 2+4+8+16+32+...-1-2-4-8-16-32. The +2 and -2 cancel each other out, so does +4 and -4, +8 and -8, etc. until all that's left is -1
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This is obviously false, we still have a term in the first sequence missing.
If the RHS ends, let's say, in 2n, the LHS will end in 2n+1. This leads us that the Σ=2n+1−1, which is true.
But we cannot evaluate a limit for n→∞, because Σ would tend also to infinity.
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(2−1)(1+2+4+8+16+32+...), you just simplify it. 2(1+2+4+8+...)−1(1+2+4+8+16+...) so it is 2+4+8+16+32...−1−2−4−8−16.... 2 and −2, 4 and −4, etc. all cancel each other out until −1 remains
Since it wasLog in to reply
1+2+3+4+5+6+…∞=12−1, requires the assumption that the R.H.S. does not end. The point of infinite sequences is that they do not end. Even the proof for 9.99999…=10 requires the 9s to never end.
However, the proof forLog in to reply
Nice logic!!
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