The third video in a series about Ramanujan.This one is about Ramanujan Summation.
Here's the wikipedia page for further reading: en.wikipedia.o...
Euler-Maclaurin Formula en.wikipedia.o...
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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)
A specific example is sum(n) = 1 + 2 + 3 + ... and sum((1-n^3)/n^2) = 0 - 7/4 - 26/9 - ....
These are both divergent series but, by Dirichlet regularization, their divergent sums are sum(n) = -1/12 and sum((1-n^3)/n^2) = (1 + 2pi^2)/12.
Finally sum(n) + sum((1-n^3)/n^2) = pi^2/6 = sum(1/n^2) as expected.
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: Grandi's series 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, and agrees with Ramanujan summation. Example: 1^-s + 2^-s + 3^-s + ... = B_(s+1)/(s+1)
For s=0 we get 1+1+1+... = -1/2 and for s=-1 we get 1+2+3+4+... = -1/12.
Some may remember the numberphile video where Tony Padilla manipulates series to write 1+2+3+... in terms of Grandi's series, and got the -1/12 answer. Strictly speaking, Tony was manipulating the Riemann zeta function, then as a last step you take s=-1. Tony explains that method here: www.nottingham....
Dirichlet regularization: This takes zeta continuation one step further and can be used for series of the form sum f(n)n^-s. This is how I got an answer sum((1-n^3)/n^2) = (1 + 2pi^2)/12 in the example sum(n) + sum((1-n^3)/n^2) = pi^2/6 = sum(1/n^2).
As you can see, convergence isn't synonymous with sum. Convergence is just one method of summation out of many. But the idea is that these different methods fit together so it makes sense to call these methods sums.