Today when I was just playing around in wiki, I came across two most interesting number series-

**1.** 1-2+3-4+5-6……..

**2.** 1-1+1-1+1-1……..

The sum of first series can be found out using simple summation of series techniques.

And it’s obvious that,

S(n) = -(n/2); if n is even.

(n+1)/2; if n is odd.

But, here comes the interesting part. This will not hold for an infinite series. Or well, that apparently following proof suggests-

4S = (1-2+3-4+….) + (1-2+3-4+….) + (1-2+3-4+….) + (1-2+3-4+….)

4S = (1-2+3-4+….) + 1 + (-2+3-4+5-..) + 1 + (-2+3-4+5-..) – 1 + (3-4+5-6+….)

4S = 1 + [(1-2-2+3) + (-2+3+3-4) + (3-4-4+5) + ………]

4S = 1 + [0+0+0+0+…..]

Hence, S = 1/4 !!!

Amazed! Here is another half of story, perhaps the better-

2S = (1-2+3-4+….) + (1-2+3-4+….)

2S = 1 + (-2+3-4+5-….) – 1 + (3+4-5+6…)

2S = 0 + (-2+3) + (3-4) + (4-5) +…..

2S = 1+1-1+1-1……..

Hence,

1+1-1+1-1+1….. =1/2 🙂

So, what is this? How it is possible?

I was also surprised for a while, but on reading further, I realised, these are not a converging series. Hence, for infinite number of terms, this becomes like finding a limit. Another basic Calculus topic [and I haven’t used/re-read it in a long while].

Further convincing (read complicated) proof requires some more of forgotten stuff, which I am not comfortable with at lease now.

So, I close it. A refreshing recap of a good ol’ topic.

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