What does \(1 + 2 + 4 + 8 + 16 + 32 + ...\) equal?
You could rewrite the equation as \((2 - 1)(1 + 2 + 4 + 8 + 16 + 32 + ...)\)
This is then written as \((2 + 4 + 8 + 16 + 32 + ...) - 1 - 2 - 4 - 8 - 16 - 32 - ...\).
Which results in \(1 + 2 + 4 + 8 + 16 + 32 + ...= -1\)!!!.
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Sharky Kesa
4 years, 11 months ago
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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 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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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 \(2^n-1\) 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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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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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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these laws are not valid when we are dealing with infinite sums.
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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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Just an illusion.....,
Confine up to 5 or 6 terms and try to evaluate..
InitiallyTrolled me too..,
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Nice logic!!
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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 \(2^n\), the LHS will end in \(2^{n+1}\). This leads us that the \( \Sigma = 2^{n+1} - 1\), which is true.
But we cannot evaluate a limit for \(n \rightarrow \infty\), because \(\Sigma\) would tend also to infinity.
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However, the proof for \(1+2+3+4+5+6+\ldots\infty=\frac{-1}{12}\), 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 \(9.99999\ldots=10\) requires the \(9\)s to never end.
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Since it was \((2 - 1)(1 + 2 + 4 + 8 + 16 + 32 + ...)\), 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
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