# 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
6 years, 11 months ago

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

- 6 years, 11 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

- 6 years, 10 months ago

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

- 6 years, 6 months ago