Summation

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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Comments

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This 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 - 6 years, 8 months ago

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 - 6 years, 8 months ago

Damn, we communicated before Schol of Excellence???

Sharky Kesa - 3 years, 7 months ago

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 - 6 years, 4 months ago

Just an illusion.....,

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

InitiallyTrolled me too..,

manoj kumar - 6 years, 8 months ago

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

Nanayaranaraknas Vahdam - 6 years, 4 months ago

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

bhavya jain - 5 years, 8 months ago

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 - 5 years, 8 months ago

It has to do with the fact that you're ignoring the last term, which is infinitely large and therefore very important

A Former Brilliant Member - 3 years, 7 months ago

There is no largest term.

Sharky Kesa - 3 years, 7 months ago

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 - 6 years, 8 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 -1

Sharky Kesa - 6 years, 8 months ago

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.

Guilherme Dela Corte - 6 years, 8 months ago

@Guilherme Dela Corte – 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

Sharky Kesa - 6 years, 8 months ago

@Guilherme Dela Corte – 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.

Nanayaranaraknas Vahdam - 6 years, 4 months ago

Nice logic!!

Anshuman Singh - 6 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$