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Summation (Part Two, Method Two)

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?

Note by Sharky Kesa
3 years, 11 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. \(\Box\)

Daniel Liu - 3 years, 8 months ago

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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/Riemannseriestheorem

Ashar Tafhim - 3 years, 11 months ago

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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.

Adarsh Kumar - 3 years, 6 months ago

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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.

Yong See Foo - 3 years, 11 months ago

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Hey Yong See, you remember me from camp?

Sharky Kesa - 2 years, 7 months ago

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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.

Muhammad Shariq - 3 years, 11 months ago

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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.

Nanayaranaraknas Vahdam - 3 years, 6 months ago

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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.

Muhammad Shariq - 3 years, 11 months ago

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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.

Ashar Tafhim - 2 years, 6 months ago

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

Nanayaranaraknas Vahdam - 3 years, 6 months ago

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