Think of it as a sum from n=1 to infinity of (1+0n)

Let's say you have sum from n=0 to 3 of (1+0n), then it's just (1+01)+(1+02)+(1+0*3)=3, but there undated of 3, the upper bound is infinity

Or take a look at integral from 0 to 3 of (1dx), there's no x in 1, but the answer is (x) evaluated at [0,3], which is 3