# Summation (Part Two, Method Two)

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?

Note by Sharky Kesa
5 years, 10 months ago

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Note 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. $\Box$

- 5 years, 8 months ago

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/Riemannseriestheorem

- 5 years, 10 months ago

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.

- 5 years, 10 months ago

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.

- 5 years, 10 months ago

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

- 5 years, 6 months ago

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.

- 4 years, 5 months ago

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.

- 5 years, 6 months ago

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.

- 5 years, 10 months ago

Hey Yong See, you remember me from camp?

- 4 years, 7 months ago

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.

- 5 years, 6 months ago