Waste less time on Facebook — follow Brilliant.
×

Summation (Part 2)

To read Summation Part 1 click here

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 ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

It is an intermediary sum for the infamous \(1+2+3+4+5+6+\ldots=?\) summation. Nanayaranaraknas Vahdam · 2 years, 8 months ago

Log in to reply

It is also \(\infty\) and \(-\infty\) if you combine them in a different way. Bogdan Simeonov · 3 years ago

Log in to reply

@Bogdan Simeonov 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 Bogdan Simeonov · 3 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...