Question

Can anyone explain why \( 1+2+3+4+...= -\dfrac{1}{12}\)?

Answer 1

No, because it doesn't?

Answer 2

Do you know zeta function?

Answer 3

\[\Large\zeta(n) = \sum_{k=1}^\infty \dfrac{1}{k^n}\]But according to Wolfram Alpha, \(\zeta(-1) = -\dfrac{1}{12}\).

http://www.wolframalpha.com/input/?i=zeta%28-1%29

Answer 4

Siths is not wrong, your question is ambiguous.

Answer 5

Sorry, I know of it, but I'm not familiar with it, unfortunately.

The way I read your series was
\[1+2+3+\cdots=\sum_{n=1}^\infty n=\infty\]
Which arguably is identical to the series
\[\sum_{k=1}^\infty \frac{1}{k^n},\text{ with }n=-1~...\]

Answer 6

Well so Wolfram Alpha is wrong?

Answer 7

Not necessarily. I may be wrong in saying the two series are identical. You'll have to forgive my ignorance regarding the zeta function.