Summation

What does 1+2+4+8+16+32+...1 + 2 + 4 + 8 + 16 + 32 + ... equal?

You could rewrite the equation as (21)(1+2+4+8+16+32+...)(2 - 1)(1 + 2 + 4 + 8 + 16 + 32 + ...)

This is then written as (2+4+8+16+32+...)12481632...(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 + ...= -1!!!.

Note by Sharky Kesa
5 years, 5 months ago

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This is an infinite geometric series which has ratio r=2>1r=2>1. So this doesn't converge, and as the result we get an non-sense answer.

Muh. Amin Widyatama - 5 years, 5 months ago

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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 2n12^n-1 which approaches to infinity as nn 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

Yong See Foo - 5 years, 5 months ago

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Damn, we communicated before Schol of Excellence???

Sharky Kesa - 2 years, 4 months ago

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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.

Nanayaranaraknas Vahdam - 5 years, 1 month ago

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Just an illusion.....,

Confine up to 5 or 6 terms and try to evaluate..

InitiallyTrolled me too..,

manoj kumar - 5 years, 5 months ago

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I had a similar derivation which I am not so sure about. My first note on Summations

Nanayaranaraknas Vahdam - 5 years, 1 month ago

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these laws are not valid when we are dealing with infinite sums.

bhavya jain - 4 years, 5 months ago

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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.

Curtis Clement - 4 years, 5 months ago

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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

Brilliant Member - 2 years, 4 months ago

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There is no largest term.

Sharky Kesa - 2 years, 4 months ago

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I do not see where 1-1 came from.

EDIT: For a proof of where this summation went wrong, check my comment down here.

Guilherme Dela Corte - 5 years, 5 months ago

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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

Sharky Kesa - 5 years, 5 months ago

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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 2n2^n, the LHS will end in 2n+12^{n+1}. This leads us that the Σ=2n+11 \Sigma = 2^{n+1} - 1, which is true.

But we cannot evaluate a limit for nn \rightarrow \infty, because Σ\Sigma would tend also to infinity.

Guilherme Dela Corte - 5 years, 5 months ago

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@Guilherme Dela Corte Since it was (21)(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 +...) so it is 2+4+8+16+32...124816...2 + 4 + 8 + 16 + 32 ...- 1 - 2 - 4 - 8 - 16.... 22 and 2-2, 44 and 4-4, etc. all cancel each other out until 1-1 remains

Sharky Kesa - 5 years, 5 months ago

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@Guilherme Dela Corte However, the proof for 1+2+3+4+5+6+=1121+2+3+4+5+6+\ldots\infty=\frac{-1}{12}, 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=10 requires the 99s to never end.

Nanayaranaraknas Vahdam - 5 years, 1 month ago

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Nice logic!!

Anshuman Singh - 5 years, 5 months ago

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