There are many instances of Ramanujan’s god like works that never occurred to other mathematicians. Even if they did, no one had the audacity to write the end results without any sort of proof or derivation. One such interesting mathematical feat consists of the summation of all natural numbers, it is also said to be the Euler-Ramanujan summation.

We celebrate National Mathematics Day by taking a look into the amazing world of Srinivasa Ramanujan.

Let’s take a look at the objective of the Euler-Ramanujan. It is to find:

**S=1+2+3+4+5+…**

The sum of such a series was considered to be infinity as per common sense, and the fact that it’s a divergent series. It still seems impossible to be solved right?

*This is how it was done.*

First, let there be a sum:

**S****1****= 1-1+1-1+1-1…****.**

This sum can be solved simply by doing this,

**S****1****=1-(1-1+1-1+1…****), **

**thus, S****1****=1- S****1****, **

**so, S****1****+S****1****=2S****1****=1, **

**so we have, S****1****=½ as our result.**

Take another sum:

**S****2****=1-2+3-4+5-6…**.

We then arrange it like this:

**S****2****=1-2+3-4+5-6…**

**S****2****= 1-2+3-4+5-6…****, and add them.**

You can try this for yourself! Simply ensure that you write the digits one below the other by shifting the first digit by just one space in the second line. What you’ll eventually get is this,

**2S****2****= 1-1+1-1+1…****. **

**So you write it as, 2S****2****=S****1****= ½. Thus, S****2****= ¼**.

Now that we have these we can proceed with the main goal,

**S=1+2+3+4+5+6+…**.

Astoundingly, this is the part where Ramanujan’s magic happened. Simply subtracted S2 from S in a traditional and simple manner:

**S=** **1+2+3+4+5+6+7+8+9+… ****–S****2****= – (1–2+3–4+5–6+7–8+9–10+…****)**

This leaves us with the equation:

**S–S****2****=4+8+12+16+20+…**,

All you have to do now is take 4 common in the right hand side and place the value of S2.

**Thus, S – ¼ = 4(1+2+3+4+5+…****), ****or algebraically,**

**S – ¼ = 4S, so, – ¼ =4S-S=3S, i.e. 3S= –¼.**

Now what do we do to find S? Simply divide both sides by 3. And as astonishing and mind blowing as it may seem, Ramanujan directly wrote to Hardy without any sort of proof that he asserts that,

**1+2+3+4+5…**** =(-1/12). **

He did this in his second long letter to Hardy, he wrote:

*“Dear Sir, *

*I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich’s Infinite Series and not fall into the pitfalls of divergent series. … I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + · · · = −1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal. I dilate on this simply to convince you that you will not be able to follow my methods of proof if I indicate the lines on which I proceed in a single letter.”*

But how is this even possible?

The **sum** of all *positive numbers* turns out to be a negative fraction.

Not a whole number, not a positive number- but a negative fraction!

As astounding as it may seem, researches in string theory use this expression and deem it to be truer than ever.

*This article was contributed by Edudigm’s student Tisyagupta Pyne.*

Sources:

The Man Who Knew Infinity – Robert Kanigel