1 − 2 + 3 − 4 + · · · = 1/4

Prove me wrong.

n is the number of digits,
if n/2 belongs to Z, then the sum is -(n/2)
if n/2 doesn't belong to Z, then the sum is -(n-1)/2 + n =
(n+1)/2

then the chances to either with n -> infi are similar, so its
[(-n + n + 1)/2]/2 and OP is correct :) lol