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1 − 2 + 3 − 4 + · · · = 1/4

Name: Anonymous 2009-05-01 0:38

Prove me wrong.

Name: Anonymous 2009-05-01 0:52

Well played, Euler.

Name: Anonymous 2009-05-01 16:14

The integers are closed and 1/4 is not an integer. Therefore, no sequence of additions or subtractions of integers will result in a number that is not an integer. QED

Name: Anonymous 2009-05-01 17:55

>>3
The integers are closed under *finite* sums.

Name: Anonymous 2009-05-01 18:19

>>4
What does it matter?

Name: Anonymous 2009-05-01 19:55

A simple test for alternating series shows it diverges.

Name: Anonymous 2009-05-01 20:09

(n=0)Σ(∞) (-1)^n * (n) is an alternating series that describes 1 - 2 + 3 - 4 + ...
The alternating series will converge to some number if the 'An' terms go to zero as n goes to infinity, in this case 'An' = n

lim n -> ∞  n = ∞

'An' goes to infinity...

...and at this point I can't remember if that means the series diverges for sure (which I think it does), or if it means a different test is required.

Name: Anonymous 2009-05-01 20:32

Cauchy criterion.

Choose epsilon < 1 and let n > N(epsilon). Then |s_n - s_(n+1)| = n+1 > 1 > epsilon.

Name: Anonymous 2009-05-01 23:06

>>1
limit test says it diverges. nice trolling though

Name: Anonymous 2009-05-01 23:56

>>5
It matters because infinite sums of integers can evaluate to non-integer values.

>>6>>7
There is no special divergence test for alt. series.

Name: Anonymous 2009-05-02 5:05

>>10
Are you claiming that you can't use induction on infinite sums?

Name: Anonymous 2009-05-02 11:09

>>4,10
You're retarded. The fact that the integers are closed precludes any sum not being an integer.

Name: Anonymous 2009-05-02 11:11

>>12

Closed *in a complete metric space* is what matters.

Name: Anonymous 2009-05-02 11:17

Who the fuck would use any sort of test to show this converges, it just really obviously doesn't.

Stop applying shit blindly and actually think about the problem for a second.

Name: Anonymous 2009-05-02 11:32

>Stop applying shit blindly and actually think about the problem for a second.

Exactly

Name: Anonymous 2009-05-02 11:57

Isn't this one of hte formulas that Ramanujan sent to Hardy while he was trying to get recognition? It took Hardy a while to discover that it was in fact a special case of the zeta function.

Name: Anonymous 2009-05-02 18:33

>>11
Yes.

For instance, you can prove by induction that 1+1/2+1/4+...+1/2^n is less than 2 for any n, but the INFINITE sum 1+1/2+1/4+... is NOT less than 2.

>>12
1+1+1+1+1+1...+1 (n "1"s) is an integer for any n, but 1+1+1+1.... is infinity, which is not an integer.

Another, possibly less stupid, example: The rationals are closed under addition, so any partial sum of 1-1/3+1/5-1/7... is a rational number, but the series sums to pi/4, which isn't rational.

>>14
>it just really obviously doesn't
you fail @ math.

>>16
Ramanujan did a lot of weird stuff that turned out to be correct, but that he didn't really justify properly (and a lot of stuff that turned out to be just wrong).

This isn't the zeta function, though (because of the signs).  From the zeta function you get 1+2+3+4+... = -1/12 since the sum is equal to the series \zeta(s)=\Sigma 1/n^s evaluated at -1. Although the series converges only for Re(s) >= 1, s != 1, the analytic continuation of the function defined by this series is holomorphic for any s != 1, and it happens that \zeta(-1)=-1/12.

Name: Anonymous 2009-05-02 19:00

-INFITIY'D

Name: Anonymous 2009-05-02 19:04

>>18
poster here.
fuck that just blew my mind when i gave it a second thought.
does not converge.

Name: Anonymous 2009-05-02 19:06

>>19
poster here.
alternates betwix - and + infinity fuck fucking fuck

Name: Anonymous 2009-05-03 16:15

>>17
I don't fail at maths, I'm three years into a university course.

That one just is obvious, the partial sums aren't even bounded.

If he meant to indicate a different sort of equality, ie. that the Cesàro sum of the series was 1/4, or defining some sort of analytic continuation of a function then he should have used appropriate notation, or at least defined his notation.

Name: Anonymous 2009-05-03 17:08

>>21
"Equality" is context-dependent.  OP didn't even say he was talking about real numbers for that matter.  He could've been talking about equality in the field Z/(1) (where everything equals zero) and he'd be perfectly correct.

Even if we stipulate that we're talking about real numbers, there's the question of how to define the sum of a series.  In the "usual" definition of convergence, the series diverges, but there's no reason to assume that was what OP was talking about, especially since it would be such an idiotic statement to begin with.

Name: Anonymous 2009-05-03 19:02

I understood the problem to be in the vein of "this obviously incorrect statement is surprisingly hard to formally disprove." or something.

Name: Anonymous 2009-05-03 21:05

the sum 1-2+3-4+... is equal to

\sum_{n = 1}^{\infty} [n*(-1)^{n+1}]

which is the same as

\sum_{n = 1}^{\infty} [(2n-1) - (2n)]

(expand it to verify it's the same series) but this reduces to

\sum_{n = 1}^{infty} (-1)

point made

Name: Anonymous 2009-05-03 22:56

>>24
interestingly you can also write an as [(2n+1) - 2n] and show that the series is +infinity. this makes perfect sense since it really is alternating between + and - infinity.

Name: Anonymous 2009-05-05 15:02

n is the number of digits,
if n/2 belongs to Z, then the sum is -(n/2)
if n/2 doesn't belong to Z, then the sum is -(n-1)/2 + n =
(n+1)/2

then the chances to either with n -> infi are similar, so its
[(-n + n + 1)/2]/2 and OP is correct :) lol

Name: Anonymous 2009-05-05 23:38

I showed this problem to my friend's little sister (9 years old). She told me the answer was 1. I'm going to have to go with what she says, because she's pretty smart.

Name: Anonymous 2009-05-06 2:13

LOL it's not even possible to add and subtract your way to a fraction with integers, not even the sum of an infinite series where each term is an integer.
Habeeb it.
QED

Name: Anonymous 2009-05-06 4:02

>>28
This seems obvious, but how would you prove it formally?

Name: Anonymous 2009-05-06 8:52

>>29
I'm not >>28, but I'd try it this way:

Let z_n be a series of integers and s_n its sequence of partial sums.

Assuming z_n was convergent, it must also pass the cauchy test, which delivers for n > N(epsilon) with epsilon < 1:

|s_n - s_(n+1)| = z_(n+1) < 1, but the only number in the integers < 1 is 0.

So that means we only have a finite amount of numbers > 0 in a convergent series of integers and with that knowledge we can resort to the closure of integers under addition.

q.e.d.

Name: Anonymous 2009-05-06 8:57

>>30
>we only have a finite amount of numbers != 0 in a convergent series of integers
Fix'd.

Negative numbers are allowed as well, of course.

Name: Anonymous 2009-05-06 9:07

>>30
>the only positive number in the integers < 1 is 0.
Fix'd.

Obvious by the context, but I might as well do it properly.

Name: Anonymous 2009-05-06 20:02

>>29
Easy. The integers are closed, so any sum of integers has to be an integer, so therefore 1/4 must be an integer, which is a contradiction.

In other news, OP is an excellent troll.

Name: Anonymous 2009-05-06 21:13

Are you still arguing about this shit? /sci/ I expected more of you :(

Name: Anonymous 2009-05-06 21:19

>>33
>any sum of integers has to be an integer
Take a look at >>17

Name: Anonymous 2009-05-06 22:08

>>35 see >>13

Name: Anonymous 2009-05-06 22:14

>>36
>any sum of integers has to be an integer
1+2+3+4+... = ?

Name: Anonymous 2009-05-07 10:56

It is 1/4.

Let's say we have n number of elements. Now you have two possibilites:

1.) 2n/2 is an integral, (n = 2 or 4, 6 and so on)
then the left side of the equation can be put as a number of sums, example: (1-2) + (3-4) + ... + [n - ( n - 1 ) ] + ...

there are n/2 sums like this one, each gives the result '-1', this gives (-1)*(n/2)=-n/2, as the left side of the equation.

2.) second possibility is that 2n/2 is NOT an integral. Then the left side of the equation is equal to:
(1-2) + (3-4) + ... + [ n - ( n + 1 ) ... + n.
in the end I add 'n' since 'n' is an odd number,
this sums up to [(n/2)*(-1) + n] = -n/2 + 2n/2 = n/2

So I have two possibilites, assuming n->infinity.

Both possibilites are equally probable, therefore truth is the average value of the sum, which is:

[ (-n/2) + n/2 ]* (1/2) = 1/4.

"ok"

Name: Anonymous 2009-05-07 10:57

>>38
then the left side of the equation can be put as a number of sums, example: (1-2) + (3-4) + ... + [n - ( n + 1 ) ] + ...

small error - it's
"+ [n - (n + 1) ] +...." of course.

Name: Anonymous 2009-05-07 12:31

>>38
Nice try, but I won't fall for this.

Name: Anonymous 2009-05-07 12:49

>>38
 is >>26
except its well more put out

Name: Anonymous 2009-05-07 12:57

>>38
"this sums up to [(n/2)*(-1) + n] = -n/2 + 2n/2 = n/2"
it does not and this is not correct, in fact it is (-1)*(n-1)*(1/2) + n; take is we take m/2 that is an integer, then if n/2 doesnt belong to integer, the closest one that does is m = n - 1;
then it looks like (m/2)*(-1) + n = -m/2 + 2n/2 = (2n - m)/2 = (2n - n +1)/2 = (n+1)/2 and there, with either possibilties having equal possibilites we take the average value, which is
[(-n/2) + (n +1)/2]/2 = 1/2/2 = 1/4;
-n + n is 0, so wheres your point? how could it possibly give 1/4 ? :)

Name: Anonymous 2009-05-07 13:32

>>6


Maybe THINK about what the problem is saying instead of blindly applying some "test" that you learned in class whose proof you probably didn't even understand.

If you think this series diverges you're crazy. Every time a number is added a another number is subtracted. It has to balance out eventually (to 1/4), like the yin and yang.

Name: Anonymous 2009-05-07 14:26

>>43
lolitrollu.jpg

Name: Anonymous 2009-05-13 11:14

>>42
YHBT

Name: Anonymous 2009-05-13 18:20

>>29
"Habeeb it" is a formal proof.
Habeeb it.

Name: Anonymous 2009-05-13 20:47

>>46
I've thought about handing in a homework problem that gives a proof along the lines of:  "You wouldn't have asked the question if you didn't have a proof, therefore a proof exists, therefore the proposition is true. QED"
Don't change these.
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