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 $1 + 2 + 4 + 8 + 16 + 32 + ...= -1$ !!!.

#CosinesGroup #Summation #NumberFallacy

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

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}`	$\boxed{123}$

Comments

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 - 7 years, 5 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 - 7 years, 5 months ago

Damn, we communicated before Schol of Excellence???

Sharky Kesa - 4 years, 4 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 - 7 years ago

Just an illusion.....,

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

InitiallyTrolled me too..,

manoj kumar - 7 years, 5 months ago

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

Nanayaranaraknas Vahdam - 7 years ago

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

bhavya jain - 6 years, 5 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 - 6 years, 5 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 - 4 years, 4 months ago

There is no largest term.

Sharky Kesa - 4 years, 4 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 - 7 years, 5 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 - 7 years, 5 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 - 7 years, 5 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 - 7 years, 5 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 - 7 years ago

Nice logic!!

Anshuman Singh - 7 years, 5 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}`	$\boxed{123}$