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Let $x = 1+1+1+1+1+1 ...$

Let $y=1-1+1-1+1-1 . . .$

First, let's find the value of $y$.

The partial sums of $y$ are $s_n=(1,0,1,0,1,0,...)$

If you take the means of the partial sums, you will get the sequence $(\frac11,\frac12,\frac23,\frac24,\frac35,...)$ which obviously converges to $\frac12$.

Another way (This time algebraically):

$$y=1-1+1-1+1-1 . . .$$

$$1-y=1-(1-1+1-1+1-1+1-1 ...)$$

$$1-y=y$$

$$2y=1$$

$$y=\frac12$$

The value of $y$ is therefore $\frac12$

$$x-y= (1+1+1+1+1+1+1 . . .)-(1-1+1-1+1-1+1-1 . . .) $$ $$x-\frac12 = (1+1+1+1+1+1+1 . . .)-(1-1+1-1+1-1+1-1 . . .) $$

$$x=(1+1+1+1+1+1+1 . . .)$$

$$-y=-(1-1+1-1+1-1+1 . . .)$$

So $x-y=$

$$(1+1+1+1+1+1+1 . . .)$$

$$(-1+1-1+1-1+1-1 . . .)$$

If you add those together, you'll see that the $-1$'s cancel out very other $1$, and that the $+1$'s add together, which equals:

$$0+2+0+2+0+2+0+2 ...$$ Or,

$$2+2+2+2+2+2+2 ...=2(1+1+1+1+1+1+1)=2x$$ Therefore, $x-y=2x$

$y=\frac12$

$x-\frac12=2x$

Subtract $x$ from both sides, and you get,

$$-\frac12=x$$

Is this a correct proof that $1+1+1+1+1+1 ...=-\frac12$?

Bretsky
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    Can it be ever true? ( I guess, No) – Inquisitive Mar 29 '15 at 04:40
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    I will never get used to Cesàro Summation, there are a lot of pathologies as (this one) you can deduce with it. – Daniel Mar 29 '15 at 04:41
  • Why can you take $y=\frac{1}{2}$? – Inquisitive Mar 29 '15 at 04:43
  • I defined that right at the start of the question. $y=1-1+1-1+1-1...=\frac12$ – Bretsky Mar 29 '15 at 04:44
  • You can't just define things to be whatever you want. Then the logic may not be consistent. You can turn these ideas into rigorous notions, but not by just willy-nilly applying "normal" algebraic manipulations to these objects you have just pulled out of a hat. The fact is $1-1+1-1\ldots$ does not converge to $\frac 1 2$, but you are manipulating it as if it did. – Matthew Levy Mar 29 '15 at 04:45
  • I don't understand what you mean by defining things to be whatever I want. How is assigning $y=1-1+1-1+1-1...=\frac12$ "pulling an object out of a hat? You may have a point that I cannot apply "normal" algebra to infinite series, which I was worried could be a problem. – Bretsky Mar 29 '15 at 04:50
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    Maybe $1-1+1-1+1-1+\ldots = e$ – Matthew Levy Mar 29 '15 at 04:52
  • While it doesn't have a sum in the typical sense, Cesàro summation can be used (along with many forms of algebraic trickery that further reinforce the Cesàro summation) to assign a value of $\frac12$ to Grandi's series. – Bretsky Mar 29 '15 at 04:59
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    Why down vote for asking a question and showing the work behind it, even if it is wrong it is a question with effort put into it +1 from me – jimjim Mar 29 '15 at 05:20
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    When you play with infinity you are attempting to be GOD. Than everything is possible, even infinity exhibiting finite quality. Your proof is correct but does not have a practical value. Just teaches us being careful how we play with math. – Moti Mar 29 '15 at 05:29
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    What OP did is simply to calculate $\zeta(0)$, though idea of regularization needs justification. – Sangchul Lee Mar 29 '15 at 05:35
  • Neither $x$ nor $y$ exist. Anything depending on them is nonsense. – MPW Mar 29 '15 at 05:37
  • How can they not exist? – Bretsky Mar 29 '15 at 05:45
  • @Bretsky: Because they are limits which do not exist. Obviously. Duh. – MPW Mar 29 '15 at 06:09

1 Answers1

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$$1+1+1+\cdots$$ will always diverge to infinity. You cannot do simple series manipulations, such as reordering or adding 0, with divergent series if you are looking for an analytical value. The proper way to do this is with Zeta regularization.

Teoc
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  • So what you are saying is that my methods were incorrect and I received the correct answer by chance, and that solving infinite series algebraically will not always give the correct solution? I never did reorder the series, as I know that could drastically change the results, but I did add $0$s. Is that why you can/should not solve questions like this algebraically? – Bretsky Mar 29 '15 at 05:42
  • @Bretsky consider this: $$S=1+1+1+1+1+\cdots$$ $$S=0+1+1+1+1+\cdots$$ $$S-S=1+0+0+0+0+\cdots$$ $$0=1$$ That is what happens. – Teoc Mar 29 '15 at 16:42
  • @Bretsky also consider this: $$S=1+1+1+1+1+...$$ $$S=1+;;;+1+;;;+1\cdots$$ $$S-S=0+1+0+1+0+1+0+\cdots$$ $$0=S$$ – Teoc Mar 29 '15 at 16:44