RAMANUJANS SEQUENCE:-

The infamous sequence goes as the following ——-

1 + 2 + 3 + 4………+ till infinity = -1/12

Now, at first glance, this might seem impossible. After all, how can the infinite sum of positive numbers return a negative number as the answer? Everyone I showed this to, and even I when I saw this for the first time, had the same reaction – that it’s a joke. However, this is very much true and is a puzzle today studied by thousands of mathematicians across the world for several decades now. The above sequence is called Ramanujan’s sequence after the mathematician Srinivasa Ramanujan, which we will study in this article.

Firstly, however, we must study two other sequences to get to the one given above. They are:-

1. 1 – 1 + 1……. = ½
2. 1-2+3-4+5……. = ¼

The above two sequences, by themselves, are extremely strange in nature. Notice how in both, the addition and subtraction of infinite integers give us a fraction as the answer, which in by itself should be impossible. However, we will now see that this is not so.

Let A = 1-1+1-1……

• -A = -1+1-1+1……

Now add 1 on both sides.

We get, 1-A = 1-1+1-1+1….. = A

• 1 = 2A
• A = ½

Now let us take the second sequence as:-

B = 1-2+3-4……

Therefore, A-B = (1-1+1-1+……) – ( 1-2+3-4………)

• A-B = (1-1) + (-1+2) + (1-3)……..

        = 1-2+3-4………

• A – B = B
• A = 2B
• B = A/2
• B = ¼

Now that we have proved the two preliminary sequences to hold true, we can now use the proved sequences to prove Ramanujan’s sequence.

Let C = 1 + 2 + 3 + 4…….. till infinity

So, B – C = (1-2+3-4……..) – (1 + 2 + 3 + 4……….)

Or, B – C = (1-1) + ( -2 – 2) + (3 – 3) + ( -4 – 4) +………

Or, B – C = (-4)(1 + 2 + 3 + 4……………..)

Or, B – C = -4C

Or, B = -3C

Or, C = B/-3 = -1/12

Hence the paradoxical sequence, which at first glance would seem to be untrue (or even on other glances as well) holds true as proved by the use of two other sequences.