The sum of all integers numbers are -1/12. Step 1: First we need to find the infinite sum of +1 - 1 + 1 ... -1 +1 ... Calling this sum S, we have: S = +1 - 1 + 1 ... -1 +1 ... Then 1-S will be 1 - S = 1 - (+ 1 - 1 + 1 ... -1 + 1 ...) , Taking off the parenthesis 1 - S = 1 - 1 + 1 - 1 ... +1 - 1 ... , because the sum is infinite we have S in the right hand side(RHS) too, so 1 - S = S, then S = 1/2 Step 2: We gonna need to the result of the sum of 1 - 2 + 3 - 4 + 5 - 6 ... Calling this sum S1, we have: S1 = 1 - 2 + 3 - 4 + 5 - 6 ..., Taking S1 + S1 we have in a special way 2S1 = 1 - 2 + 3 - 4 + 5 - 6 ... + 0 + 1 - 2 + 3 - 4 + 5 - 6 ...., this is a smart way to rearrange the sum. Now we sum term by term the first line with the second(Like 0 + 1 / -2 + 1/ +3 - 2/ -4 + 3/ ...) Then we gonna have 2S1 = 1 - 1 + 1 - 1 + 1 - 1 ...., that we already now the result of RHS 2S1 = 1/2, so S1 = 1/4 Step 3: Now we can actually prove 1 + 2 + 3 + 4 + ... is equal to -1/12. Calling this sum S2, we have S2 = 1 + 2 + 3 + 4 + ..., taking off S1, we have S2 - S1 = 1 + 2 + 3 + 4 + ... - ( 1 - 2 + 3 - 4 + 5 - 6 + ...) S2 - S1 = 1 + 2 + 3 + 4 + ... - 1 + 2 - 3 + 4 - 5 + 6 - ... S2 - S1 = 1 + 2 + 3 + 4 + ... - 1 + 2 - 3 + 4 - 5 + 6 - ..., Making the sum term by term of the first with the second (Like 1 -1 / 2 + 2 / 3 - 3/ ...) S2 - S1 = 4 + 8 + 12 + 16 + ...., the right hand side is 4S2, so S2 - S1 = 4S2, we know that S1=1/4, so 3S2 = -1/4 S2 = -1/12 haha