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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 ) .

letputun=∑k=1nk(k!)1)provethatun=(n+1)!−12)studytheconvergenceof∑n=1∞1un.

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 .

1)letprovethisrelationbyrecurrencep(1)u1=2!−1=1×(1!)p(1)istrueletsuposeun=(n+1)!−1⇒un+1=∑k=1n+1k(k!)=∑k=1nk(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)!−1theimplicationp(n)⇒p(n+1)istrue.2)wehave∑n=1∞1un=∑n=1∞1(n+1)!−1butforn→∞1(n+1)!−1∼1(n+1)!andΣ1(n+1)!isconvergentsoΣ1uniscovergent.

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

un=∑nk=1(k+1−1)(k!)un=∑nk=1[(k+1)!−k!]Ittelescopesandbecomesun=(n+1)!−1

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