To read Method One click here.

What is $1 - 2 + 3 - 4 + 5 - ...$ equal?

Let's call the sum $s$.

$2s = (1 - 2 + 3 - 4 + ...) + (1 - 2 + 3 - 4 + ...)$

$2s = 1 + (- 2 + 3 - 4 + 5 - ...) + 1 - 2 + (3 - 4 + 5 - 6 + ...)$

$2s = 0 + (- 2 + 3) + (3 - 4) + (- 4 + 5) +...$

$2s = 1 - 1 + 1 - 1 + 1 ...$

$2s = 1 + (- 1 + 1) + (- 1 + 1) + ...$

$2s = 1 + 0 + 0 + 0 + ...$

$s = \frac {1}{2}$

But, from my previous method, the answer was $\frac {1}{4}$!

How is this possible?

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## Comments

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TopNewestNote that $1-1+1-1+\cdots=\dfrac{1}{2}$, via another famous summation. Therefore $2s=\dfrac{1}{2}$, and $s=\dfrac{1}{4}$ as desired. $\Box$

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This is because of Rienmann series thereom. This sequence is conditionally convergent so rienmann series theorem states that it could be rearranged in order to get any number. http://en.wikipedia.org/wiki/Riemann

seriestheoremLog in to reply

What you have done here is mathematically incorrect. You can't pair up terms like that. As a matter of fact on the line $2s=1-1+1-1+1+...+(-1)^n$ is a geometric series with $r=-1$ and the sum diverges, i.e. it does not sum up to a specific value (A geometric series only converges when $|r|<1$). The correct method to solve this problem is to pair up the terms as:

$s = 1-2+3-4+5...=\lim_{n \rightarrow \infty}(1+3+5+...+(2n+1))-(2+4+6+...+(2n))=$

$\lim_{n \rightarrow \infty}(n^2)-(n^2+n)=\lim_{n \rightarrow \infty}-n=-\infty$.

As you can see the sum diverges (tends to $-\infty$) and so is undefined.

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I'd like to add that the way your summing the series, it is even possible to get 1/8/, 1/16....and so on as the answer which is clearly absurd.

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However, this series yields a finite sum, which is clearly defined, because of a unique property of infinity.

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yes it is possible . you should search rienmann series theorem which states that if a series is conditionally convergent it can always be re arranged to get any number.

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The series of $1-1+1-1+1\ldots$ is the infamous Grandi series, which yields the sums $0, 1$ and $\frac12$. However, for practical purposes, $\frac12$ is the accepted value, as $1$ and $0$ can be yielded by just changing the parentheses.

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Again, using the same method in your previous link that I mentioned, the partial sums diverge, so the sum also diverges. The partial sums for the first 2k terms is -1, for the first 2k+1 terms is (k+1)/2. Clearly the partial sums diverge.

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Hey Yong See, you remember me from camp?

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There are two possibilities:1)the one shown in the note.2)the,series of( 1s )(-1s) ending with a -1 then the answer would be 0.

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