Summation (Part 2)

To read Summation Part 1 click here

What is 1−2+3−4+5−6+...1 - 2 + 3 - 4 + 5 - 6 + ... equal?

Let's call the sum ss.

s=(1−2+3−4...)s= (1 - 2 + 3 - 4...)

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

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

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

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

4s=1+[(1−2−2+3)+(−2+3+3−4)+(3−4−4+5)+(−4+5+5−6)+...]4s= 1 + [(1 - 2 - 2 + 3) + (- 2 + 3 + 3 - 4) + (3 - 4 - 4 + 5) + (- 4 + 5 + 5 - 6) + ...]

4s=1+[0+0+0+0+...]4s= 1 + [0 + 0 + 0 + 0 + ...]

4s=14s= 1

s=14s= \frac {1}{4}!!!

#Summation #NumberFallacy

Note by Sharky Kesa
7 years, 5 months ago

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1 vote

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Comments

It is also ∞\infty and −∞-\infty if you combine them in a different way.

Bogdan Simeonov - 7 years, 5 months ago

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I've written a post about this sum. This is equal to η(−1) \eta(-1) and is factually equal to a quarter, as you said! It is not just a fallacy. See eta function here

Bogdan Simeonov - 7 years, 5 months ago

It is an intermediary sum for the infamous 1+2+3+4+5+6+…=?1+2+3+4+5+6+\ldots=? summation.

Nanayaranaraknas Vahdam - 7 years ago
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