Determine the sum of ∑n=2∞1n!\sum_{n=2}^{\infty} \frac{1}{n!}∑n=2∞n!1 given e=∑n=0∞1n!e = \sum_{n=0}^{\infty} \frac{1}{n!}e=∑n=0∞n!1.
e−2e-2e−2
e−1e-1e−1
eee
e−1.5e-1.5e−1.5