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Un-k-1-n-1-n-k-show-that-the-sequence-converges-and-determine-the-limit-

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Question Number 201070 by Rodier97 last updated on 29/Nov/23
Un = Σ_(k=1) ^n (1/ ((n),(k) )) show that the sequence converges and determine the limit
$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Un}\:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}} \\ $$$$ \\ $$$${show}\:\:{that}\:{the}\:{sequence}\:{converges}\:{and} \\ $$$${determine}\:{the}\:{limit}\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by Frix last updated on 29/Nov/23
(1/ ((n),(k) ))=((k!(n−k)!)/(n!)) U_n =1+Σ_(k=1) ^(n−1) ((k!(n−k)!)/(n!)) =1+(1/(n!))Σ_(k=1) ^(n−1) k!(n−k)! The greatest summands are those with k=1, k=n−1. Their sum is (2/n). The next are k=2, k=n−2. Their sum is (4/(n^2 −n)) k=3, k=n−3 ⇒ ((12)/(n^3 −3n^2 +2n)) ... (2/n) (2/n)+(4/(n^2 −n))=((2n+2)/(n^2 −n)) (2/n)+(4/(n^2 −n))+((12)/(n^3 −3n^2 +2n))=((2n^2 −2n+8)/(n^3 −3n^2 +2n)) ... We end up with ((an^p +...)/(n^(p+1) +...)) lim_(n→∞) ((an^p +...)/(n^(p+1) +...)) =0 ⇒ lim_(n→∞) U_n =1
$$\frac{\mathrm{1}}{\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}}=\frac{{k}!\left({n}−{k}\right)!}{{n}!} \\ $$$${U}_{{n}} =\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\:\frac{{k}!\left({n}−{k}\right)!}{{n}!}\:=\mathrm{1}+\frac{\mathrm{1}}{{n}!}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\:{k}!\left({n}−{k}\right)! \\ $$$$\mathrm{The}\:\mathrm{greatest}\:\mathrm{summands}\:\mathrm{are}\:\mathrm{those}\:\mathrm{with} \\ $$$${k}=\mathrm{1},\:{k}={n}−\mathrm{1}.\:\mathrm{Their}\:\mathrm{sum}\:\mathrm{is}\:\frac{\mathrm{2}}{{n}}.\:\mathrm{The}\:\mathrm{next} \\ $$$$\mathrm{are}\:{k}=\mathrm{2},\:{k}={n}−\mathrm{2}.\:\mathrm{Their}\:\mathrm{sum}\:\mathrm{is}\:\frac{\mathrm{4}}{{n}^{\mathrm{2}} −{n}} \\ $$$${k}=\mathrm{3},\:{k}={n}−\mathrm{3}\:\Rightarrow\:\frac{\mathrm{12}}{{n}^{\mathrm{3}} −\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}} \\ $$$$… \\ $$$$\frac{\mathrm{2}}{{n}} \\ $$$$\frac{\mathrm{2}}{{n}}+\frac{\mathrm{4}}{{n}^{\mathrm{2}} −{n}}=\frac{\mathrm{2}{n}+\mathrm{2}}{{n}^{\mathrm{2}} −{n}} \\ $$$$\frac{\mathrm{2}}{{n}}+\frac{\mathrm{4}}{{n}^{\mathrm{2}} −{n}}+\frac{\mathrm{12}}{{n}^{\mathrm{3}} −\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}}=\frac{\mathrm{2}{n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{8}}{{n}^{\mathrm{3}} −\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}} \\ $$$$… \\ $$$$\mathrm{We}\:\mathrm{end}\:\mathrm{up}\:\mathrm{with}\:\frac{{an}^{{p}} +…}{{n}^{{p}+\mathrm{1}} +…} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{an}^{{p}} +…}{{n}^{{p}+\mathrm{1}} +…}\:=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{U}_{{n}} \:=\mathrm{1} \\ $$

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