Seems legitimate to me

Algebra Level 3

Line 1: 1 + 2 + 3 + . . . + n = n ( n + 1 ) 2 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}

Line 2: n = 1 n = 1 12 \displaystyle \sum_{n=1}^{\infty} n = \frac{-1}{12} (Considering Riemann zeta regularisation)

Line 3: (When n = n = \infty ): 1 12 = ( + 1 ) 2 \frac{-1}{12} = \frac{\infty(\infty+1)}{2}

Line 4: 2 + + 1 6 = 0 \infty^2 + \infty + \frac{1}{6} = 0

Line 5: = ( 1 ) ± ( 1 ) 2 4 ( 1 ) ( 1 6 ) 2 ( 1 ) \infty = \frac{-(1) \pm \sqrt{(1)^2 - 4(1)(\frac{1}{6})}}{2(1)}

Line 6: = 3 ± 3 6 \infty = \frac{-3 \pm \sqrt{3}}{6}

Where does this proof first go wrong?

Line 4 (You can't expand the brackets and rearrange like that) Line 3 (You can't equate the formulae) Line 1 (That isn't a valid formula) Line 5 (The quadratic formula substitution is incorrect) There is no error, and we need to rewrite the laws ofmathematics Line 2 (That isn't a valid formula) Line 6 (There is an error in the manipulation)

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