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# Summation (Part 2)

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

Let's call the sum $$s$$.

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

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

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

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

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

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

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

$$4s= 1$$

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

Note by Sharky Kesa
3 years, 4 months ago

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

It is also $$\infty$$ and $$-\infty$$ if you combine them in a different way. · 3 years, 4 months ago

I've written a post about this sum. This is equal to $$\eta(-1)$$ and is factually equal to a quarter, as you said! It is not just a fallacy. See eta function here · 3 years, 4 months ago