# From the Simpsons to Riemann’s paradox: discussing Riemann’s rearrangement theorem.

What if I told you, you could subtract ∞ from ∞ to get π, i.e, ∞–∞ = π? Isn’t that absolutely absurd? This is the Riemann’s paradox which was discovered nearly 150 years ago! And today we are going to talk a little about Riemann’s paradox and the Riemann rearrangement theorem. But before we begin on that, lets first talk a little about the last equation in the Simpson’s black board, which is the expansion of the natural log of 2:

To begin with, we take a few more terms,

Now we rearrange a bit, adding the firs and second term, the third and sixth, the fifth and tenth and so on, leaving the terms which lie in a serial number multiple of 4,

Simplifying, we get,

This is worrisome, if you look closely, the terms of equation (2) which is derived directly from equation (1) are half of that of equation (1), or this indirectly says 1 = 2. Well, math cannot be broken so you might think we must be making mistake here, however we have not! This is one of the basic ideas of infinite series, they hold true only when considered for a given particular order and for any other order their prediction stops making sense. This is the Riemann’s rearrangement theorem, that says, manipulating the order in given infinite series we can end up with any desirable result, or in other words the commutative theorem doesn’t exactly work for infinite series.

In fact for the example of the convergent series of ln 2, considered above, by Riemann’s rearrangement theorem, manipulations can be made to make the series result into π or even e! Interesting, right? However, let’s just take it at face value for now, as reaching ~3.14 would take summing at least a couple 100 terms, if summing serially. Instead, let us do something to the infinite series itself into something absolutely crazy!

First, we separate the negative and the positive terms of the series, so:

Now, lets see if the sum of positive terms is convergent or divergent,

As the denominators in all of the cases are odd numbers, we can write the sum of the positive part of the infinite series as

We test, if the series is convergent or divergent using the integral test, as f(n) here 1/(2n+3) is both positive and decreasing,

The integral test states that for ∑ of f(n) from 0 -> ∞, where f(n) fits the criteria, if ∫ of f(x)dx from 0 to ∞ converges, that is, is equal to a value, then the initial summation converges as well and vice versa

So, evaluating the following integral we get

Now, as b tends to ∞, ln (2b+3) tends to ∞ as well, thus

Thus, the sum of the positive terms of the series is divergent or in other words when summed over infinite terms will not lead to a constant but to ∞. Similarly, it can also be proved using the same method that the sum of the negative numbers is also divergent and equals -∞. When putting, keeping Riemann’s rearrangement theorem in mind, this into the equation that we started with, we get:

Mind blowing, isn’t it? In fact, infinitely many similar equations that make absolutely no sense at all, all using the Riemann’s rearrangement theorem.

So, remember to keep the terms of your infinite series in order or they might just get crazy! Homer says, “thank you for reading”