1 + 2 + 3 + 4 + ... Equals ... -1/12?!Date: 09/18/2012 at 14:07:21 From: sankar Subject: divergent series I have read on the Internet about Ramanujan's divergent series, where 1 + 2 + 3 + 4 + ..... = -1/12 How is this possible? Date: 09/23/2012 at 21:14:37 From: Doctor Vogler Subject: Re: divergent series Hi Sankar, Thanks for writing to Dr. Math. This is a common point of confusion. Allow me to explain what's going on here. Recall that in order for a sequence to converge, the terms must get smaller and smaller and go to zero. But the converse is NOT true: Just because the terms go to zero does NOT imply that the series converges. For example, it turns out that this series is very close to (within 1 of) the natural logarithm of n (that is, ln n): 1 + 1/2 + 1/3 + 1/4 + ... + 1/n It increases more and more slowly as n gets bigger, but it keeps increasing, and ln n grows to infinity. So you are correct that this grows to infinity, or diverges: 1 + 2 + 3 + ... Now, it seems pretty bizarre to suggest that it could be a negative fraction, namely -1/12. But it turns out that this function can be extended in a unique way -- known as analytic continuation -- to all complex numbers except s = 1 (where the function grows to infinity): zeta(s) = 1/1^s + 1/2^s + 1/3^s + 1/4^s + ... This converges for s > 1, and for all complex numbers s the real part of which is bigger than 1. And *this* function has the value -1/12 when s = -1. See also http://mathforum.org/library/drmath/view/52831.html This is much like summing the series ... 1 + x + x^2 + x^3 + x^4 + ... ... and getting ... 1/(1 - x) ... and then evaluating this at x = 2 to get 1 + 2 + 4 + 8 + 16 + ... = -1. It's ludicrous, because the series doesn't converge at x = 2. But it converges (for x < 1) to a function that *does* have a value when x = 2. That's essentially what happens with analytic continuation. (Indeed, if you treat x as a complex number, then analytic continuation extends the geometric series to the fraction I mentioned.) For come more information, you could have a look at http://en.wikipedia.org/wiki/Analytic_continuation This, too: http://en.wikipedia.org/wiki/Riemann_zeta_function If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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