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Ramanujan Summation

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• published: 01 May 2016
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The third video in a series about Ramanujan.This one is about Ramanujan Summation. Here's the wikipedia page for further reading: https://en.wikipedia.org/wiki/Ramanujan_summation Euler-Maclaurin Formula https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula --------- Here is an example of divergent summation, that hopefully shows its usefulness, and shows that it is rigorous and consistent with traditional summation. Imagine a series c_n that can be split into two other series a_n and b_n as follows: sum c_n = sum (a_n + b_n) = sum a_n + sum b_n Using this we can work out sum c_n from the values of sum a_n and sum b_n. This is standard stuff when working with finite series, or convergent series. It is also possible for a convergent series to be the sum of two divergent series. So in the above, c_n is convergent and a_n, b_n are divergent. In that case, we can still work out sum c_n from the values of sum a_n and sum b_n, but now you have to use divergent summation. This only works if the divergent summation method is regular (gives the same answer as convergent summation when applied to c_n) and linear (so sum (a_n + b_n) = sum a_n + sum b_n) Some series are are harder to sum than others. So there are levels of summation that you can use. The levels are something like this: Finite series: Can be added together, multiplied, rearranged, as expected. Convergent series: Has all the properties of finite series, except a sequence of partial sums does not end with the value of the series. Instead, the limit of the sequence is used as the sum of the series. Example: geometric series with decreasing terms 1 + 1/2 + 1/4 + ... = 2 Conditionally convergent series: Has all the properties of convergent series, but if you rearrange the terms you get different answers. Example: 1 - 1/2 + 1/3 - 1/4 + .... = ln(2) Divergent series: Would go to infinity by definition of convergent series. Various other methods can be applied to give a value. Some divergent series are harder to give a value than others. See below. Any divergent summation methods needs to agree with the limit when applied to convergent series (i.e. regular). Divergent series Cesaro summation: Can still be added and multiplied like convergent series (i.e. linear). Example: 1 - 1 + 1 - 1 + ... = 1/2 Divergent series Euler summation: A method of analytic continuation. Still can be added and multiplied as expected (linear). Example: geometric series with increasing terms 1 + 2 + 4 + 8 +... = -1 Divergent series Borel summation: Can give a value to harder series but still agrees with previous methods. Loses a property known as stability, where removing a term from the series does not simply subtract its value from the total sum. Adding and multiplying (linearity) still exists. Divergent series Ramanujan summation: Still linear. Can be used on the most stubborn divergent series, but depends on your choice of a parameter. I called the parameter 'a' in the video. When a is taken as infinity this method agrees with convergent sums. Examples when a=0: 1+1+1+1+... = -1/2. 1+2+3+4+...=-1/12. Example when a=1: 1 + 1/2 + 1/3 + 1/4 + ... = 0.5772... the euler-mascheroni constant. Zeta Function continuation: A method of analytic continuation. You will continuously approach the value, but the method is nonlinear. Agrees with Ramanujan summation. Example: 1^s + 2^s + 3^s + ... = B_(s+1)/(s+1) As you can see, convergence isn't synonymous with sum. Convergence is just one method of summation out of many.