# Sums That Add to Zero, the Gateway to the Riemann Hypothesis

**Episode #5 of the course Great math problems for the 21st-century mind by John Robin**

To lead us into the true meat of the Riemann Hypothesis, I am going to claim something that you will probably not believe:

1 + 2 + 3 + 4 + … = −1/12

Sound weird? Get ready! This is only the start of the strangeness we’ll encounter as we extend our understanding of the p-series.

First of all, let’s talk about what we expect this sum to be. Infinity, right?

Indeed, to our common senses, adding any series of whole numbers should add to infinity. For example:

1 + 4 + 9 + 16 + 25 + …

or

1 + 8 + 27 + 64 + 125 + …

or more generally,

1 + 2^{n} + 3^{n} + 4^{n} + 5^{n} + …

should all add to infinity, right?

Before we explore that, do you notice something? You’ll remember from yesterday’s lesson on the p-series that we showed that for

1 + 1/(2^{n}) + 1/(3^{n}) + 1/(4^{n}) + 1/(5^{n}) + …

the series converges for all n greater than 1. I also told you that there were other possibilities.

Here’s that thing I wanted you to notice:

1 + 2 + 3 + 4 + 5 + … = 1 + 1/(2^{–1}) + 1/(3^{–1}) + 1/(4^{–1}) + 1/(5^{–1}) + …

More generally,

1 + 2^{n} + 3^{n} + 4^{n} + 5^{n} + … = 1 + 1/(2^{–n}) + 1/(3^{–n}) + 1/(4^{–n}) + 1/(5^{–n}) + …

My opening examples, then, are cases of the p-series for all negative integer numbers.

Now, back to my original claim, which we can now state as: The p-series converges to −1/12 for n = −1.

To prove this, I am going to create two sums.

S_{1} = 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + …

and

S_{2} = 1 − 2 + 3 − 4 + 5 − 6 + 7 − …

and we will call our original sum S—i.e., S = 1 + 2 + 3 + 4 + …

Now, for S_{1}, notice that the value will alternate between 0 and 1. If you stop at an even term (say the fourth term, so 1 − 1 + 1 − 1), you will get 0. If you stop at an odd term (say the fifth term, so 1 − 1 + 1 − 1 + 1), you will get 1. We let the “sum” of S_{1} then be the average, so 1/2.

To find S_{2}, we need to do a little trick. We will write S_{2} twice and arrange the terms in a convenient way:

2S_{2} = 1 − 2 + 3 − 4 + 5 − 6 + …

1 − 2 + 3 − 4 + 5 − …

and we’ll then add up each term down the column to get:

2S_{2} = 1 − 1 + 1 − 1 + 1 − 1 + …

Notice this is exactly S_{1}.

This means we know:

2S_{2} = S_{1}

And since we know S_{1} = 1/2, that means 2S_{2} = 1/2, and therefore S_{2} = 1/4.

Now, with this, I will prove my claim by using a similar trick:

S − S_{2} = 1 + 2 + 3 + 4 + 5 + 6 + …

−[1 − 2 + 3 − 4 + 5 + 6 − …]

S − S_{2} = 1 + 2 + 3 + 4 + 5 + 6 + …

−1 + 2 − 3 + 4 − 5 + 6 − …

Adding up each term down the column, we get:

S − S_{2} = 0 + 4 + 0 + 8 + 0 + 12 + …

= 4[1 + 2 + 3 + … ]

= 4S

Applying that S_{2} is 1/4, we get:

S − 1/4 = 4S

which simplifies to

−1/4 = 3S

which further simplifies to

S = −1/12

i.e., 1 + 2 + 3 + 4 + 5 + 6 + … = −1/12, and in other words, the p-series for n = −1 converges to −1/12.

Though this makes no sense to our intuition, in other settings, like those of string theory, it makes a lot of sense. In fact, it makes a lot of sense in these settings that 1 + 1/(2^{n}) + 1/(3^{n}) + 1/(4^{n}) + 1/(5^{n}) + … converges always to a finite number for all negative integer values of n.

Furthermore, for all even values (n = 2, n = 4, etc.), this sum converges to exactly 0.

Yes, that’s right:

1 + 4 + 9 + 16 + 25 + … = 0

and

1 + 16 + 81 + 256 + 625 + … = 0

and

1 + 2^{2n} + 3^{2n} + 4^{2n} + 5^{2n} + … = 0

Understanding the Riemann Hypothesis is all about understanding what’s happening here to make the value zero. Tomorrow, we will explore this further, by way of expanding to the complex numbers and defining the Riemann zeta function.

Stay tuned!

**Recommended video**

Proof That 1 + 2 + 3 + … = −1/12 by Numberphile

**Recommended book**

*The Elegant Universe* by Brian Greene

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