Summation (Part Two, Method Two)

To read Method One click here.

What is 12+34+5...1 - 2 + 3 - 4 + 5 - ... equal?

Let's call the sum ss.

2s=(12+34+...)+(12+34+...)2s = (1 - 2 + 3 - 4 + ...) + (1 - 2 + 3 - 4 + ...)

2s=1+(2+34+5...)+12+(34+56+...)2s = 1 + (- 2 + 3 - 4 + 5 - ...) + 1 - 2 + (3 - 4 + 5 - 6 + ...)

2s=0+(2+3)+(34)+(4+5)+...2s = 0 + (- 2 + 3) + (3 - 4) + (- 4 + 5) +...

2s=11+11+1...2s = 1 - 1 + 1 - 1 + 1 ...

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

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

s=12s = \frac {1}{2}

But, from my previous method, the answer was 14\frac {1}{4}!

How is this possible?

Note by Sharky Kesa
5 years, 10 months ago

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1 vote

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Note that 11+11+=121-1+1-1+\cdots=\dfrac{1}{2}, via another famous summation. Therefore 2s=122s=\dfrac{1}{2}, and s=14s=\dfrac{1}{4} as desired. \Box

Daniel Liu - 5 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 - 5 years, 10 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=11+11+1+...+(1)n2s=1-1+1-1+1+...+(-1)^n is a geometric series with r=1r=-1 and the sum diverges, i.e. it does not sum up to a specific value (A geometric series only converges when r<1|r|<1). The correct method to solve this problem is to pair up the terms as:

s=12+34+5...=limn(1+3+5+...+(2n+1))(2+4+6+...+(2n))=s = 1-2+3-4+5...=\lim_{n \rightarrow \infty}(1+3+5+...+(2n+1))-(2+4+6+...+(2n))=

limn(n2)(n2+n)=limnn=\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 - 5 years, 10 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 - 5 years, 10 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 - 5 years, 6 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 - 4 years, 5 months ago

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The series of 11+11+11-1+1-1+1\ldots is the infamous Grandi series, which yields the sums 0,10, 1 and 12\frac12. However, for practical purposes, 12\frac12 is the accepted value, as 11 and 00 can be yielded by just changing the parentheses.

Nanayaranaraknas Vahdam - 5 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 - 5 years, 10 months ago

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Hey Yong See, you remember me from camp?

Sharky Kesa - 4 years, 7 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 - 5 years, 6 months ago

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