# Why is the sum of 1-2+3-4+... is 1/4?

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• Apr 24th 2010, 06:20 AM
chengbin
Why is the sum of 1-2+3-4+... is 1/4?
Can someone explain to me the paradox 1-2+3-4+... =1/4? I really don't get why and how. Thanks so much.
• Apr 24th 2010, 06:38 AM
Opalg
Quote:

Originally Posted by chengbin
Can someone explain to me the paradox 1-2+3-4+... =1/4? I really don't get why and how. Thanks so much.

The binomial series $\displaystyle 1-2x+3x^2-4x^3+\ldots$ converges to $\displaystyle (1+x)^{-2}$ when $\displaystyle |x|<1$. If you ignore that restriction on x and pretend that the series also converges to the same function when x=1 then you come up with the formula $\displaystyle 1-2+3-4+\ldots = 1/4$, which of course is not true in any conventional sense.
• Apr 24th 2010, 07:26 PM
Call $\displaystyle S:= 1-1+1-\cdots$. Cancelling off the very first term by subtracting 1 gives $\displaystyle -1+1-1+\cdots= - (1-1+1-\cdots)$ so $\displaystyle S = -(S - 1)$ or $\displaystyle S=\frac12$.

Call $\displaystyle T := 1-2+3-4+\cdots$. Add the series $\displaystyle 1-2+3-4+\cdots$ and $\displaystyle 1-1+1-1+\cdots$ together term by term to get $\displaystyle T+S = (1+1) - (2+1) + (3+1) - \cdots = 2 - 3 + 4 -\cdots = -(T-1)$. Since $\displaystyle S=\frac12$, $\displaystyle T+\frac12 = -(T-1)$ or $\displaystyle T=\frac14$.

The remarkable thing, of course, is that you get the same answer if you do it "algebraically" like this as if you do it like Opalg does it (analytically, generating funtions).
• Apr 24th 2010, 07:35 PM
Prove It
Quote:

Originally Posted by chengbin
Can someone explain to me the paradox 1-2+3-4+... =1/4? I really don't get why and how. Thanks so much.

$\displaystyle 1-2+3-4+\dots = (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + \dots$

$\displaystyle = (-1) + (-1) + (-1) + (-1) + \dots$

$\displaystyle = \lim_{n \to \infty}\frac{n}{2}(-1)$.

This is clearly not $\displaystyle \frac{1}{4}$...
• Apr 24th 2010, 08:05 PM
lvleph
There is no way this is 1/4. The reason? Because, integers are closed under addition.
• Apr 24th 2010, 08:34 PM
Any convergent series whose terms are integers has only finitely many non-zero terms. But closure only applies to finite sums. For instance, the rationals are closed under addition but the sum $\displaystyle 1+\frac1{2!}+\frac1{3!}+\frac1{4!}+\cdots$ is irrational.
• Apr 25th 2010, 05:44 AM
lvleph
Looks like I need to read up on my Real Analysis again, because this is not making sense to me.
• Apr 25th 2010, 05:56 AM
Bacterius
Lvleph is right in a sense. Since you only consider the addition and the substraction of integer terms, you cannot possibly end up with a non-integer result. So the original statement cannot be right !
• Apr 25th 2010, 06:20 AM
chengbin

1 ? 2 + 3 ? 4 + · · · - Wikipedia, the free encyclopedia

I don't know enough math to understand the "proofs" in that article. Can anyone explain the number manipulation method? It seems like the proof that requires the least amount of math knowledge to understand.
• Apr 25th 2010, 07:39 AM
lvleph
Quote:

Originally Posted by Bacterius
Lvleph is right in a sense. Since you only consider the addition and the substraction of integer terms, you cannot possibly end up with a non-integer result. So the original statement cannot be right !

Things get weird when we begin talking infinity though. But then again, the abstract algebra sense of the Integers does include an infinite collection. Even the linear algebra sense gives you closure since the Integers are a vector space.

EDIT: I can see how this would be 1/4 in the formal power sense (which is what maddas was saying), but this sub forum is not that advanced.
• Apr 25th 2010, 08:32 AM
Wilmer
Nothing else to do, so I'm trying:
1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 -10 - 11 + 12 + 13 ....

Well, this gives me running totals of 1, -4, 5, -8, 9, -12, 13 ...

Now how in heck can I paradoxize this (Nerd)
• Apr 25th 2010, 08:54 AM
tonio
Quote:

Originally Posted by lvleph
There is no way this is 1/4. The reason? Because, integers are closed under addition.

This is a very bad reason: also rationals are closed under addition yet $\displaystyle \sum^\infty_{k=0}\frac{1}{n!}=e\notin\mathbb{Q}$ ...(Giggle)

Tonio
• Apr 25th 2010, 08:57 AM
tonio
Quote:

Originally Posted by Bacterius
Lvleph is right in a sense. Since you only consider the addition and the substraction of integer terms, you cannot possibly end up with a non-integer result. So the original statement cannot be right !

Nop, this is not a good reason: the conterexample $\displaystyle \sum^\infty_{k=0}\frac{1}{n!}=e\notin\mathbb{Q}$ shows that a sum of only rational numbers can have a non-rational sum...of course, as already noted, the gist here is that the sum is infinite and thus closedness and other related things do not apply here.

Tonio
• Apr 25th 2010, 09:00 AM
tonio
Quote:

Originally Posted by lvleph
Things get weird when we begin talking infinity though. But then again, the abstract algebra sense of the Integers does include an infinite collection. Even the linear algebra sense gives you closure since the Integers are a vector space.

They are? Under what field, say?

Tonio

EDIT: I can see how this would be 1/4 in the formal power sense (which is what maddas was saying), but this sub forum is not that advanced.

.
• Apr 25th 2010, 09:18 AM