Question

Question #ae3c8

Answer 1

`e/2`

Explanation:

We have

`(1+1/n)^n approx 1 + n/n+(n(n-1))/(2! n^2)+(n(n-1)(n-2))/(3! n^3) + cdots +`

and

`e = 1 + 1 + 1/(2!)+1/(3!) + cdots +`

then

`e -(1+1/n)^n approx sum_(k = 2)^m(1/(k!)-(prod_(j=0)^(k-1) (n-j))/(k! n^k)) = sum_(k = 2)^m1/(k!)(1-(prod_(j=0)^(k-1) (n-j))/(n^k))`

and

`lim_(n->oo)n(1-(prod_(j=0)^(k-1) (n-j))/(n^k)) = ( (k-1)k)/2` and then

`lim_(n->oo)n(e -(1+1/n)^n) = 1/2sum_(k=2)^oo ((k-1)k)/(k!) = e/2`