Question #ae3c8

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Answer 1 · 2017-11-13T14:55:39

$e/2$

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

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