Debunking Ramanujan's Finite Solutions to Divergent Series

ABSTRACT:

This paper shows how an infinite series yields both an infinite solution and a finite solution and how to determine which one is correct and which one is just plain crazy!

Imagine the following geometric series:

Is s convergent or divergent? Does it blow up to infinity or have a finite value? So far, we can't say. However, according to Ramanujan and his apologists, even if s is divergent, it has a finite value. Let's see if this is really the case. With a little algebra we derive equation 5 below:

What solutions will work for equation 5? Here are two:

One solution is infinity. If k is finite, the other solution is finite. This means, if we want, we can always extract a finite solution from a divergent series. It also means we can always extract an infinite solution from a convergent series! Ramanujan claimed that divergent series have finite solutions. He could have (but didn't) claim that convergent series have infinite solutions. If such claims seem counter-intuitive and just plain crazy, it's because they are. Extracting finite solutions from divergent series (or infinity from convergent series) violates the geometric series rules. Here are the rules:

At 9 we have a system of equations that show that the value of x matters. We don't just get to choose willy-nilly which equation we like best. If we choose the appropriate equation for a convergent series, we get a finite solution as we should. If we choose the appropriate equation for a divergent series, we get infinity as we should when x is greater than or equal to 1. Albeit, our system of equations doesn't cover x when it is equal to or less than -1. Let's start with the case where x equals -1:

Equation 10 doesn't converge, nor does it blow up to infinity. The partial sums oscillate between 1 and 0 forever. Mathematical physicists like to take the average of .5. However, it is obvious there are two solutions: 1 and 0. In the case where x is less than -1, the partial sums oscillate toward infinity and minus infinity. Again, there are two solutions. For negative numbers less than or equal to -1 we need two equations:

The complete system of equations is as follows:

Where x is less than or equal to -1, we can take advantage of a peculiar property of infinity: Infinity minus infinity doesn't have to equal zero, since infinity equals any number plus infinity:

Thus when x is less than -1 we can take the average of the two solutions s1 and s2 and get a finite solution. In fact, since n can be any finite number, there can be an infinite number of finite solutions.

Where x equals -1, there is one finite solution: .5.

Now, given what we know about infinity, finite numbers and the geometric series rules, what kind of solution(s) can we extract from the following series:

This is a tricky one, since there is no x variable to tell us which equation to use. So let's go with the usual method. We first take the average of the "a" series:

Then we do some fancy Ramanujan algebra to get the average of the "b" series:

Now that we have the b-series average, we can find solutions to the c-series:

At 38 and 39 above, we have two solutions: infinity and -1/12. One solution seems correct and the other seems counter-intuitive. Are they both valid? There is a test: simply sum up the series. The fastest way to do that is find the average number in the series, then multiply it by the number of terms. The average number and the number of terms are both infinity:

As expected, c = infinity. For any series, a good rule of thumb is if the series is divergent, a finite solution is invalid and just plain crazy! We can tell if a series equals infinity by looking at the last term. If the last term is infinite, then the series sum is infinite because infinity plus a bunch of other numbers equals infinity. We get a finite solution because the series itself does not know if it is convergent or divergent. Both convergent and divergent series are structured the same way: a sum of numbers--so the algebra is the same for both and yields more than one solution. Thus it is up to us to take a further step and test each solution.