Sum of an Infinite Arithmetic Series

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SUMMARY

The sum of the infinite arithmetic series $$\sum_{n=1}^{\infty}n$$ is represented by the Riemann Zeta function, specifically \zeta(-1), which equals -1/12. This result is derived through a mathematical technique known as analytic continuation, which allows for the evaluation of functions beyond their initial convergence limits. It is crucial to understand that this does not imply the series converges in the traditional sense; rather, it is a representation that leads to a counterintuitive negative value.

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  • Understanding of infinite series and convergence
  • Familiarity with the Riemann Zeta function
  • Knowledge of complex analysis, particularly analytic continuation
  • Basic mathematical notation and operations involving series
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  • Study the properties and applications of the Riemann Zeta function
  • Learn about analytic continuation in complex analysis
  • Explore the implications of divergent series in mathematics
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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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