Can anyone tell the proof of
1+2+3+4+5+6+7+8+........upto infinity= -1/12
?

Question

Ethan Adkins · Accepted Answer

Noone can proove this thesis, because it is false.

The left side of the expression is a sum of an infinite sequence. Only geometrical sequence can have finite sum of all terms, but this sequence is not a geometrical one therfore it is not convergent (i.e. does not have finite sum).

Eva Adamson · Answer

Here's a "proof" from Srinivasa Ramanujan...

The most straightforward non-rigorous "proof" is credited to Srinivasa Ramanujan and goes something like this: #c = 1+2+3+4+5+6+7+8+...# #4c=0+4+0+8+0+12+0+16+...# After subtracting, we obtain: #-3c = 1-2+3-4+5-6+7-8+...# Now: #1/(1+x)^2 = 1-2x+3x^2-4x^3+5x^4-6x^5+7x^6-8x^7+...# So putting #x=1# we find: #-3c = 1-2+3-4+5-6+7-8+... = 1/(1+1)^2 = 1/4# Then dividing both ends by #-3# we get: #c = -1/12# Keep in mind that manipulating divergent infinite series in this manner is not really valid. The calculations above are a shadow of the real derivation of the Ramanujan Sum of the series #1+2+3+4+...#, which is more properly presented using the Riemann Zeta function and analytic continuation. The useful thing about the above non-rigorous derivation is that it gives a very rough sketch of the direction of the proper one. Ramanujan developed techniques that are used in quantum field theories to formally assign finite values to divergent infinite sums.

Nathan Axel · Answer

undefined The statement that $1 + 2 + 3 + 4 + \ldots$ sums to $-\frac{1}{12}$ is not correct. The series $1 + 2 + 3 + 4 + \ldots$ diverges to positive infinity. However, in certain contexts like string theory and analytic continuation, this divergent series can be assigned a value of $-\frac{1}{12}$ for specific purposes, but it's crucial to understand that this is not the same as saying the sum converges to $-\frac{1}{12}$.

Can anyone tell the proof of 1+2+3+4+5+6+7+8+........upto infinity= -1/12 ?