MHB Sum of an Infinite Arithmetic Series

AI Thread Summary
The sum of the infinite arithmetic series, represented as ∑n from 1 to infinity, is often stated to equal -1/12, which can be confusing since it implies a negative value for the sum of natural numbers. This result arises from the Riemann Zeta function, specifically ζ(-1), which does not converge in the traditional sense. The concept of analytic continuation allows for the calculation of ζ(-1) in a way that yields -1/12, although this does not represent a conventional sum. It is important to note that this technique is not universally applicable to all mathematical expressions. The discussion highlights the complexities and misconceptions surrounding infinite series and their sums.
IHateFactorial
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Somewhere I saw that the sum of the infinite arithmetic series

$$\sum_{n=1}^{\infty}n = \frac{-1}{12}$$

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.
 
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IHateFactorial said:
Somewhere I saw that the sum of the infinite arithmetic series

$$\sum_{n=1}^{\infty}n = \frac{-1}{12}$$

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 [math]\sum_{n = 1}^{\infty} n [/math] is a representation of something called the Riemann Zeta function, [math]\zeta (-1)[/math]. 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 [math]\zeta (-1)[/math] can be calculated and comes out to -1/12. (The form of the zeta function given by analytic continuation in this domain is not [math]\sum_{n = 1}^{\infty} n[/math] 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
 
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