ABSTRACT
Letting n→∞, we obtain e ≤ s. On the other hand, if n > m, then
(1 + 1/n)n > 2 + m∑ k=2
1 k! (1− 1/n)(1− 2/n) · · · (1− (k − 1)/n).
Letting n→∞, we see that e ≥ sm. Letting m→∞ yields e ≥ s. ♦
arn, where a, r ∈ R and a 6= 0, converges iff |r| < 1, in which case
arn = a1− r .
