∞∑n=1C(n,0)+C(n,1)+...+C(n,n)P(n,n)==ek+m.Find k+m ?
We have ∞∑n=1C(n,0)+C(n,1)+...+C(n,n)P(n,n)
=∞∑n=1nC0+nC1+...+nCnn!
=∞∑n=12nn!
=21!+222!+233!+...
=e2−1