n→∞lim(1+n1)n=j=0∑∞j!1
Apply binomial expansion on the left hand side, we have:
n→∞lim(1+n1)n=n→∞lim(1+1!n⋅n1+2!n(n−1)n21+3!n(n−1)(n−2)n31+⋯)=n→∞lim(1+1!1nn+2!1n2n(n−1)+3!1n3n(n−1)(n−2)+⋯)=n→∞lim(1+1!1(1)+2!1(1)(1−n1)+3!1(1)(1−n1)(1−n2)+⋯)=1+1!1+2!1+3!1+⋯=0!1+1!1+2!1+3!1+⋯=j=0∑∞j!1