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In asking a question about analysis textbooks there was a bit of a chat about things like S = 1 - 1 + 1 - 1 ... = 1 - (1 + -1 +1 -1 ...) or 2S = 1 or S=1/2. I will say straight away the answer to what's going on - what infinite sums are, are simply definitions and believe it or not there are a number of them - but students usually only learn one. However they all have some general properties and those general properties allow us to often calculate the sum even without knowing what definition we are using.

Well we got onto the good old Ramanujan Sum 1 +2 + 3 + 4 ... = -1/12. Believe it or not as I will show in a you tube video link its used in string theory and elsewhere in physics. I will give my proof which is a little different to others.

Define C(s) = 1^s + 2^s + 3^s ... We get the sum we want when s=1. It turns out using standard methods of complex analysis (analytic continuation and what not) it exists when s=1. So how to calculate it:

2*2^s*C(s) = 2*2^s + 2*4^s +2*6^s ...

(1 - 2*2^s)*C(s) = 1^s - 2^s + 3^s - 4^s ...

Let s=1 so you have -3*C(1) = 1 - 2 + 3 - 4 ...

Now 1 - 2 +3 - 4 ... = 1 - (1 - 1 + 1 - 1 ...) - (1 - 2 + 3 - 4 ...)

So -3*C(1) = 1 - 1/2 + 3*C(1) or C(1) = 1 + 2 + 3 + 4 ... = -1/12

Wow - it should be infinite - or should it? Whats going on - answer - it depends on your definition of infinite sum.

Now have a look at the following videos:

And if you want more heavy math see the blog by Terry Tao the guy that started me thinking about this:

https://terrytao.wordpress.com/2010...tion-and-real-variable-analytic-continuation/

Now the 64 million dollar question is this, Look in video 1 - he opens a string theory text - low and behold - we use the -1/12 result. But this is not the only area - see:

https://en.wikipedia.org/wiki/Zeta_function_regularization

What's going on here. Sure in physics you want finite answers - but why this funny definition of summation?

Thanks

Bill

Well we got onto the good old Ramanujan Sum 1 +2 + 3 + 4 ... = -1/12. Believe it or not as I will show in a you tube video link its used in string theory and elsewhere in physics. I will give my proof which is a little different to others.

Define C(s) = 1^s + 2^s + 3^s ... We get the sum we want when s=1. It turns out using standard methods of complex analysis (analytic continuation and what not) it exists when s=1. So how to calculate it:

2*2^s*C(s) = 2*2^s + 2*4^s +2*6^s ...

(1 - 2*2^s)*C(s) = 1^s - 2^s + 3^s - 4^s ...

Let s=1 so you have -3*C(1) = 1 - 2 + 3 - 4 ...

Now 1 - 2 +3 - 4 ... = 1 - (1 - 1 + 1 - 1 ...) - (1 - 2 + 3 - 4 ...)

So -3*C(1) = 1 - 1/2 + 3*C(1) or C(1) = 1 + 2 + 3 + 4 ... = -1/12

Wow - it should be infinite - or should it? Whats going on - answer - it depends on your definition of infinite sum.

Now have a look at the following videos:

And if you want more heavy math see the blog by Terry Tao the guy that started me thinking about this:

https://terrytao.wordpress.com/2010...tion-and-real-variable-analytic-continuation/

Now the 64 million dollar question is this, Look in video 1 - he opens a string theory text - low and behold - we use the -1/12 result. But this is not the only area - see:

https://en.wikipedia.org/wiki/Zeta_function_regularization

What's going on here. Sure in physics you want finite answers - but why this funny definition of summation?

Thanks

Bill

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