lim_(n→∞) Σ_(k=1) ^n ((k!(n−k)!)/(n!)) = 1


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Question Number 127454 by snipers237 last updated on 29/Dec/20

$\lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{k! (n - k)!}{n!} = 1$
	Answered by Ar Brandon last updated on 29/Dec/20
	$∀n≥4, ∀k∈{2,...,n−2}, ^n C_k ≥((n(n−1))/2) u_n =Σ_(k=1) ^n ((k!(n−k)!)/(n!))=Σ_(k=0) ^n (1/( ^n C_k ))−1 ={(1/( ^n C_0 ))+(1/( ^n C_1 ))+Σ_(k=2) ^(n−2) (1/( ^n C_k ))+(1/( ^n C_(n−1) ))+(1/( ^n C_n ))}−1={2+(2/n)+Σ_(k=2) ^(n−2) (1/( ^n C_k ))}−1 ⇒u_n +1≤2+(2/n)+Σ_(k=2) ^(n−2) (2/(n(n−1)))=2+(2/n)+((2(n−3))/(n(n−1))) ⇒2+(2/n)≤u_n +1≤2+(2/n)+((2(n−3))/(n(n−1))) ⇒lim_(n→∞) (u_n +1)=2 ⇒ lim_(n→∞) u_n =1$
	$\forall n ⩾ 4, \forall k \in {2, . . ., n - 2}, \overset{n}{} C_{k} ⩾ \frac{n (n - 1)}{2}$ $u_{n} = \underset{k = 1}{\sum^{n}} \frac{k! (n - k)!}{n!} = \underset{k = 0}{\sum^{n}} \frac{1}{\overset{n}{} C_{k}} - 1$ $= {\frac{1}{\overset{n}{} C_{0}} + \frac{1}{\overset{n}{} C_{1}} + \underset{k = 2}{\sum^{n - 2}} \frac{1}{\overset{n}{} C_{k}} + \frac{1}{\overset{n}{} C_{n - 1}} + \frac{1}{\overset{n}{} C_{n}}} - 1 = {2 + \frac{2}{n} + \underset{k = 2}{\sum^{n - 2}} \frac{1}{\overset{n}{} C_{k}}} - 1$ $\Rightarrow u_{n} + 1 ⩽ 2 + \frac{2}{n} + \underset{k = 2}{\sum^{n - 2}} \frac{2}{n (n - 1)} = 2 + \frac{2}{n} + \frac{2 (n - 3)}{n (n - 1)}$ $\Rightarrow 2 + \frac{2}{n} ⩽ u_{n} + 1 ⩽ 2 + \frac{2}{n} + \frac{2 (n - 3)}{n (n - 1)}$ $Double subscripts: use braces to clarify$
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