Sum of an Infinite Arithmetic Series

[IHateFactorial]

Somewhere I saw that the sum of the infinite arithmetic series

\(\displaystyle \sum_{n=1}^{\infty}n = \frac{-1}{12}\)

Why exactly is this? I thought infinite arithmetic series had no solution? Also... WHY is it negative? Seems counter-intuitive that the sum of all the NATURAL numbers is a decimal, a negative decimal.

 

[topsquark]

Somewhere I saw that the sum of the infinite arithmetic series

\(\displaystyle \sum_{n=1}^{\infty}n = \frac{-1}{12}\)

Why exactly is this? I thought infinite arithmetic series had no solution? Also... WHY is it negative? Seems counter-intuitive that the sum of all the NATURAL numbers is a decimal, a negative decimal.

This is a popular topic and there is a lot of misinformation out there about it. As stated the LHS does not exist and certainly wouldn't add up to a fraction on the RHS, much less as a negative value.

This goes a bit deep and can be very confusing. The sum \(\displaystyle \sum_{n = 1}^{\infty} n \) is a representation of something called the Riemann Zeta function, \(\displaystyle \zeta (-1)\). The confusion arises because this form of the zeta function cited here does not converge. BUT we can use a technique called "analytic continuation" in the complex plane and write the zeta function in a form where \(\displaystyle \zeta (-1)\) can be calculated and comes out to -1/12. (The form of the zeta function given by analytic continuation in this domain is not \(\displaystyle \sum_{n = 1}^{\infty} n\) so we aren't taking an infinite sum.)

This might seem a bit like magic and I need to stress that you cannot do this for just any expression. Analytic continuation does not always increase the size of the domain that you can calculate a function over...the zeta function just happens to be one of them that you can do this with.

-Dan