Do you know the sum of 1 to infinity is -1/12🤓🤓😮😮😮🤔🤔🤯

           Have you ever imagined 

What will be the sum of               1+2+3+4+5+.........+∞

Do you know the formula 

n(n+1)/2 for finding sum of first 'n' natural number. 

Example: if we to find sum of  first 20 natural numbers then,we apply the above formula

n(n+1/)2 

20(20+1)/2

(20*21)/2

=210

But in case of infinity

We don't know what will have to place in place of 'n' in formula "n(n+1)/2"

But a great Indian Mathematician Mr. Srinivas Ramanujan solves this problem, and  prove that sum of 1 to  ∞ is = -1/12

Then we know the equation 1+2+3+4+5+.........+∞ = -1/12 as Ramanujan Summation.

Let's see the proof

Let

A = 1-1+1-1+1-1+1-1................∞

       We expect that after solving the equation we get A= 0

But our supposition is wrong, let's see why👇👇👇👇

1-A = 1-(1-1+1-1+1-1+1-1................∞)

       After opening the bracket the symbols got changed 👇👇👇👇👇

1-A = 1-1+1-1+1-1+1-1+1.................∞

1-A = A

1=A+A

1=2A

A= 1/2

        It is amazing to get that 

1-1+1-1+1-1+1-1.........∞ = 1/2

Let's do an another equation

B= 1-2+3-4+5-6+7...............∞

A-B = (1-1+1-1+1-1+1-1+.....)-(1-2+3-4+5-6+7.........)

      After opening the bracket, all symbols  in brackets get changed 

A-B = (1-1+1-1+1-1+1-1+.....)-1+2-3+4-5+6-7.............

     Let's do a little shuffling(rearranging)

that is how, I did reshuffling

A-B = (1-1)+ (-1+2)+(1-3)+(-1+4)+(1-5)+(-1+6)+(1-7).......

A-B = 0+1-2+3-4+5-6+7..........

A-B = 1-2+3-4+5-6+7............ It is same as B

A-B = B

A= B+B

A = 2B

1/2 = 2B.           (A = 1/2) proved above

B = 1/4

This means 

1-2+3-4+5-6+7.......... = 1/4

Is this equations getting amazing, isn't it🤓🤓🤓🤓🤓🤓🤓🤓🤨😮😮😮😮

Let's do the last step

Let

C = 1+2+3+4+5+6+7............

B-C = (1-2+3-4+5-6+7.........) - (1+2+3+4+5+6+7+...............)

      All symbols in brackets got changed 👇👇👇👇👇👇👇👇👇👇👇👇

B-C = (1-2+3-4+5-6+7........)-1-2-3-4-5-6-7-.............

      Let's do a little bit shuffling again

B-C = (1-1)+(-2-2)+(3-3)+(-4-4)+(5-5)+(-6-6)+(7-7)+............

B-C = 0+(-4)+0+(-8)+0+(-12)+.........

B-C = -4-8-12....................

   Taking -4 as common

B-C = -4(1+2+3+4+5..............)

B-C = -4(C)

B-C = -4C

B = -4C+C

B = -3C

1/4 = -3C.      (B = 1/4) proved above

C = -1/12

This means

1+2+3+4+5+6+7.........∞ = -1/12

        

     Do you think, this is true

It doesn't seems to be true because sum of positive integer gives us a negative integer ,also which is in fraction

This Ramanujan Summation is used in solving many equation of physics

Let's appreciate our Indian legend  Sir 'Srinivas Ramanujan'.

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