More about infinity
Is there anything in the real world that is infinite? Not yet. We don’t have anything that is infinite. We can only talk about infinity as a process. But, we are not aware of any infinite process in the universe. Talking about infinity is like talking about some that does not exist. But, inspite of this, there are many researches and discoveries that have been proven using infinity in math.
One of the greatest mathematicians of his time, Srinivasa Ramanujan, had worked on many infinite series. One of the most interesting series was,
1 + 2 + 3 + 4 + 5 + ……… = — 1/12
After Ramanujan had found this, while sending it to G.H.Hardy, he told him “Please don’t think I am crazy, but, it’s true.” It seems impossible, doesn’t it? How can the sum of infinite positive natural numbers be a negative fraction? But, it was proved in quite a few ways and I’m going to show one of the proofs to you.
Before I prove the above equation, I’ll first tell you about another series that I will use in the proof. This is called the Grandi’s series. It goes as follows,
1–1 + 1–1 + 1–1 + 1–1 + 1–1 + ……..
Also written as,
If you stop the series at an odd place, the answer will be 1, but, if you stop at an even place, the answer will be 0. That’s obvious right? The speciality of the Grandi’s series is that it’s sum is half.
1–1 + 1–1 + 1–1 + 1–1 + 1–1 + …….. = 1/2 — (i)
Seems impossible again right? Let’s prove it.
If you write the series as (1–1) + (1–1) + (1–1) + …….. the answer will be 0. But, if you rewrite the series as 1 — ( 1–1) + (1–1) + (1–1) + ….. by removing the minus sign from the series, the answer now will be 1. Surprisingly, there is a third answer and that is 1/2.
Let G = 1–1 + 1–1 + 1–1 + ……..
Then,
1 — G = 1 — ( 1–1 + 1–1 + 1–1 + ……..)
= 1–1 + 1–1 + 1–1 + 1 — ……..
= G
Thus, 1 — G = G
1 = 2G
G = 1/2.
Well, now, let’s go back to Ramanujan’s series.
To start proving that, we are going to use result (i). Let’s call it S1.
So, S1 = 1–1 + 1–1 + 1–1 + 1–1 +……. = 1/2
Let’s take S2 such that, S2 = 1–2 + 3–4 + ….. , and,
S = 1 + 2 + 3 + 4 + 5 + …….
Now, if you take S2 and add it to itself,
2(S2) = ( 1–2 + 3–4 + 5 — ……) + ( 1–2 + 3–4 + 5 — …..)
But, while calculating this, we add the 2nd term of the first series (i.e, -2) with the 1st term of the second series (i.e, 1) and the 3rd term of the first series (i.e, 3) with the 2nd term of the second series (i.e, -2). We continue to calculate like that upto infinity. While calculating that way, our first term of the answer will be 1, (i.e, 1 + 0), the second term will be -1, (i.e, -2 + 1) and the third term will be 1, (i.e, 3–2 ) and so on. Thus we will get 2(S2) as,
1–1 + 1–1 + 1–1 + 1–1 +……. But, this is 1/2.
Thus, 2(S2) = 1/2
S2 = 1/4 — (ii)
Therefore, 1–2 + 3–4 + 5 — …… = 1/4
Let’s now calculate (S — S2)
= ( 1 + 2 + 3 + 4 + 5 + …….) — ( 1–2 + 3–4 + 5 — ……)
= 4 + 8 + 12 + 16 + …..
= 4( 1 + 2 + 3 + 4 + 5 + …….)
= 4(S)
Thus, S — S2 = 4S
We know that, S2 = 1/4 , ( from (ii) )
S — 1/4 = 4S
-1/4 = 3S
S = -1/12.
Thus, 1 + 2 + 3 + 4 + 5 + ……. = — 1/12.
Hence, proved.
Fascinating, isn’t it? Intuitively this is impossible. But s you can see, it has been proved by many and Ramanujan himself.
