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 \(1 + 2 + 4 + 8 + 16 + 32 + ...= -1\)!!!.

## Comments

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TopNewestThis 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. – Muh. Amin Widyatama · 3 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

infiniteseries! and does not end. – Nanayaranaraknas Vahdam · 3 years agoLog in to reply

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 – Yong See Foo · 3 years, 4 months ago

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– Sharky Kesa · 3 months, 3 weeks ago

Damn, we communicated before Schol of Excellence???Log in to reply

It has to do with the fact that you're ignoring the last term, which is infinitely large and therefore very important – Shubham Dhull · 3 months, 3 weeks ago

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– Sharky Kesa · 3 months, 3 weeks ago

There is no largest term.Log in to reply

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 · 2 years, 4 months ago

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these laws are not valid when we are dealing with infinite sums. – Bhavya Jain · 2 years, 4 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 · 3 years ago

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Just an illusion.....,

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

## InitiallyTrolled me too..,

– Manoj Kumar · 3 years, 4 months agoLog in to reply

Nice logic!! – Anshuman Singh · 3 years, 4 months ago

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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. – Guilherme Dela Corte · 3 years, 4 months ago

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– Sharky Kesa · 3 years, 4 months ago

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 -1Log in to reply

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. – Guilherme Dela Corte · 3 years, 4 months ago

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does notend. 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. – Nanayaranaraknas Vahdam · 3 years agoLog in to reply

– Sharky Kesa · 3 years, 4 months ago

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 \(-1\) remainsLog in to reply