Let S = 1+2+3+4+5+6+7

Consider S1= 1-1+1-1+1-1+1-1…..

Now, this sum should be 0 or 1 based on number of natural numbers taken. If infinite numbers are even, S1=0, if odd S1=1. But, Riemann zeta function gives it a value of ½. Mathematical community too agrees that the sum is ½. How? At first instance, it feels like the whole community of mathematicians are playing a prank. It is like celebrating April fools day! But yes, serious mathematical work went into the proof. If you are interested to know, please go through Ramanujan’s summation principles and zeta function. Let me try a simple proof avoiding all the complexity.

S1=1-1+1-1+1-1+1…..

1-S1=1-(1-1+1-1+1-1+1…)

1-S1=1-1+1-1+1-1+1…..

1-S1=S1

So, S1=1/2

Several objections can be put like why not the alternative solution of 0 or 1? But as stated before, we need much powerful tools in mathematics like zeta functions to come to unique solution of ½. For now, we could agree S1=1/2.

Let S2=1-2+3-4+5-6+7…..

So, S2=1-2+3-4+5-6+7-8+9…..

S2= 1-2+3-4+5-6+7-8……. I have shifted RHS by a unit position

+ 2S2=1-1+1-1+1-1+1…..

Hence, 2S2=S1

Therefore, S2=1/4

Let’s come back to our sum of infinite numbers.

S=1+2+3+4+5+6+7+8+9…..

S2=1-2+3-4+5-6+7-8+9….

So, S-S2=4+8+12+16+20…..

Hence, S-S2=4(1+2+3+4+5+6+7+8….)

S-S2=4S

So, -S2=3S

And, S=-S2/3=-1/12

If you satisfy on my answer please mark it BRAINLIEST ✏️ ✏️✏️✏️✏️✏️✏️✏️✏️✏️✏️✏️