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

Note by Sharky Kesa
3 years ago

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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. · 3 years 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. · 2 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 · 3 years 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. · 2 years ago

these laws are not valid when we are dealing with infinite sums. · 2 years ago

I had a similar derivation which I am not so sure about. My first note on Summations · 2 years, 8 months ago

Just an illusion.....,

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

# InitiallyTrolled me too..,

· 3 years ago

Nice logic!! · 3 years ago

I do not see where $$-1$$ came from.

EDIT: For a proof of where this summation went wrong, check my comment down here. · 3 years 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 · 3 years 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. · 3 years ago

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. · 2 years, 8 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 $$-1$$ remains · 3 years ago