Sum of the Natural Numbers Puzzle Solution

The solution to my most recent puzzle is that there could be two solutions. I’ll deal with the obvious solutions first. That is infinity. This is pretty obvious really. The series is clearly divergent; it doesn’t converge to a finite limit. When adding all the natural numbers up to infinity, the sum is always going to be infinite.

I say there are two solutions but the second solution only appears when you almost manipulate mathematics. You have to redefine what summation means to find this answer. The answer is -1/12. This initially seems ridiculous. You are adding up all these positive integers and you receive a negative answer. There are many ways of proving this result. I am going to explain the one Euler came up with. Ramanujan also had a very nice proof for this but his involves a very complicated formula.

The First thing Euler did was to explore the sum below.

1+ x + x² + x³ + x⁴ + x⁵…….

The sum below is a geometric progression(g.p.). The sum to infinity of a g.p. is a/1-r, where a is the first value in the sequence and r is the common ratio. This sum to infinity is true when -1<x<1. This sum to infinity tells us that the sum to infinity of the sequence above is 1/1-x. Therefore ,we have the equation below.

1+ x + x² + x³ + x⁴ + x⁵……….=1/1-x

We can then differentiate both sides. This gives us the equation below. The left hand side is differentiated using the basic method and the right and side uses the quotient rule.

1 + 2x + 3x² + 4x³ + 5x⁴ ……..= 1/(1-x)²

We can then plug in a value for x. When x=-1 we get the sequence below. This is technically not allowed because when x=-1 the sequence has no sum to infinity. Steps like these are why proofs of this result are a bit shaky. It is quite apparent when you group the sum into pairs, the sum to infinity of this sum will be negative infinity, but let’s ignore that.

1 – 2 + 3 – 4 + 5 -6 ……=1/4

We now have to bring in something called the Riemann-Zeta function. This function is massively important to mathematics, in particular it can be used to give information on the distribution of prime numbers. The Zeta function was first created by Euler and is stated below, z(s) is just the function notation for the Zeta function.

z(s) = 1^-s + 2^-s + 3^-s + 4^-s …..

Then Euler played around with the Zeta function. First he multiplied the function by s^-2, which leaves the equation below.

s^-2 x z(s) = 2^-s + 4^-s + 6^-s + 8^-s …..

 He then took away two lots of zeta multiplied by s^-2 from zeta, which produces the sum below. The terms need to aligned as in the equation below.

(1-2.s^-2) x z(s) = 1 + 2^-s + 3^-s + 4^-s + 5^-s……

                   -2(        2^-s             4^-s                …….

(1-2.s^-2) x z(s) = 1 - 2^-s + 3^-s – 4^-s + 5^-s…….

With the new equation we have just produced we are going to set s=-1. This produces the equation below.

-3 x z(-1) = 1 - 2 + 3 – 4 + 5 – 6….

The sum on the right hand side we have already proved is equal to 1/4. Therefore, we can rewrite the above as the below.

-3 x z(-1) = ¼

What is z(-1)? It is the infinite sum 1 + 2 + 3 + 4 + 5…..

Look that is the sequence of natural numbers. Therefore, we can solve for the sum of the natural numbers below.

-3( 1 + 2 + 3 + 4 + 5……)=1/4

1 + 2 + 3 + 4 + 5 + 6 ….. = -1/12

There we are proved. Well maybe. I believe the best way of thinking about this result is how Mathologer described it in his excellent video. He said the sum of the natural numbers on our planet is infinity. However, on a different planet with slightly different definitions of mathematics, the sum of the natural numbers could be -1/12.

Thank you for reading. My next puzzle will be online soon.