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
Messages
15
Reaction score
0
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.
 
Mathematics news on Phys.org
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
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top