Yes - I mostly left the first part in for the humor (it was legitimately my first stab at the problem), but it gives the same result, just in a way that’s a little harder (to me, at least) to see. The cancellation is unbalanced: Each numerator cancels with the next denominator, which necessarily brings with it the next numerator - you’ve always got the next numerator in line, as-is, after any number of canceled pairs. So while everything cancels out in the limit, the product up to n equals n+1, and so the limit of the product is ∞.
reply
Yes - I mostly left the first part in for the humor (it was legitimately my first stab at the problem), but it gives the same result, just in a way that’s a little harder (to me, at least) to see. The cancellation is unbalanced: Each numerator cancels with the next denominator, which necessarily brings with it the next numerator - you’ve always got the next numerator in line, as-is, after any number of canceled pairs. So while everything cancels out in the limit, the product up to n equals n+1, and so the limit of the product is ∞.