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Question Number 32273 by abdo imad last updated on 22/Mar/18

let put u_n =Σ_(k=1) ^n k(k!) 1) prove that u_n =(n+1)! −1 2) study the convergence of Σ_(n=1) ^∞ (1/u_n ) .

$${let}\:{put}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\left({k}!\right) \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{u}_{{n}} =\left({n}+\mathrm{1}\right)!\:−\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{convergence}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{u}_{{n}} }\:. \\ $$

Commented by abdo imad last updated on 25/Mar/18

1) let prove this relation by recurrence p(1) u_1 = 2!−1=1×(1!) p(1) is true let supose u_n =(n+1)!−1 ⇒u_(n+1) =Σ_(k=1) ^(n+1) k(k!) =Σ_(k=1) ^n k(k!) +(n+1)(n+1)! =(n+1)!−1 +(n+1)(n+1)!=(n+1)!(n+1+1) −1 =(n+2)(n+1)! −1 =(n+2)! −1 the implication p(n)⇒p(n+1) is true . 2)we have Σ_(n=1) ^∞ (1/u_n ) = Σ_(n=1) ^∞ (1/((n+1)! −1)) but for n→∞ (1/((n+1)! −1)) ∼ (1/((n+1)!)) and Σ (1/((n+1)!)) is convergent so Σ(1/u_n ) is covergent .

$$\left.\mathrm{1}\right)\:{let}\:{prove}\:{this}\:{relation}\:{by}\:{recurrence} \\ $$$${p}\left(\mathrm{1}\right)\:{u}_{\mathrm{1}} =\:\mathrm{2}!−\mathrm{1}=\mathrm{1}×\left(\mathrm{1}!\right)\:{p}\left(\mathrm{1}\right)\:{is}\:{true}\:{let}\:{supose} \\ $$$${u}_{{n}} =\left({n}+\mathrm{1}\right)!−\mathrm{1}\:\Rightarrow{u}_{{n}+\mathrm{1}} =\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:{k}\left({k}!\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} {k}\left({k}!\right)\:+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)! \\ $$$$=\left({n}+\mathrm{1}\right)!−\mathrm{1}\:+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)!=\left({n}+\mathrm{1}\right)!\left({n}+\mathrm{1}+\mathrm{1}\right)\:−\mathrm{1} \\ $$$$=\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)!\:−\mathrm{1}\:=\left({n}+\mathrm{2}\right)!\:−\mathrm{1}\:{the}\:{implication} \\ $$$${p}\left({n}\right)\Rightarrow{p}\left({n}+\mathrm{1}\right)\:{is}\:{true}\:. \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{u}_{{n}} }\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!\:−\mathrm{1}}\:{but}\:{for}\:{n}\rightarrow\infty \\ $$$$\:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!\:−\mathrm{1}}\:\sim\:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}\:{and}\:\:\Sigma\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}\:{is}\:{convergent}\:{so} \\ $$$$\Sigma\frac{\mathrm{1}}{{u}_{{n}} }\:{is}\:{covergent}\:. \\ $$

Answered by Tinkutara last updated on 23/Mar/18

u_n =Σ_(k=1) ^n (k+1−1)(k!) u_n =Σ_(k=1) ^n [(k+1)!−k!] It telescopes and becomes u_n =(n+1)!−1

$${u}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left({k}+\mathrm{1}−\mathrm{1}\right)\left({k}!\right) \\ $$$${u}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\left({k}+\mathrm{1}\right)!−{k}!\right] \\ $$$${It}\:{telescopes}\:{and}\:{becomes} \\ $$$${u}_{{n}} =\left({n}+\mathrm{1}\right)!−\mathrm{1} \\ $$

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