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Question Number 192927 by 073 last updated on 31/May/23

Answered by aba last updated on 31/May/23

n!(n+1)^m ≈(√(2πn))((n/e))^n ×n^m ∧ (n+m)!≈(√(2π(n+m)))(((n+m)/e))^(n+m) ⇒((n!(n+1)^m )/((n+m)!))≈((√(2πn))/( (√(2π(n+m)))))×(n^(n+m) /e^n )×(e^(n+m) /((n+m)^(n+m ) )) ⇒((n!(n+1)^m )/((n+m)!))≈((√n)/( (√(n+m))))×((n/(n+m)))^(n+m) ×e^m ⇒((n!(n+1)^m )/((n+m)!))≈(√(1−(m/(n+m))))×(1−(m/(n+m)))^(n+m) ×e^m lim_(n→+∞) ((n!(n+1)^m )/((n+m)!))=lim_(n→+∞) (√(1−(m/(n+m))))×(1−(m/(n+m)))^(n+m) ×e^m =1×e^(−m) ×e^m =1

n!(n+1)m≈2πn(ne)n×nm∧(n+m)!≈2π(n+m)(n+me)n+m⇒n!(n+1)m(n+m)!≈2πn2π(n+m)×nn+men×en+m(n+m)n+m⇒n!(n+1)m(n+m)!≈nn+m×(nn+m)n+m×em⇒n!(n+1)m(n+m)!≈1−mn+m×(1−mn+m)n+m×emlimn→+∞n!(n+1)m(n+m)!=limn→+∞1−mn+m×(1−mn+m)n+m×em=1×e−m×em=1

Answered by MM42 last updated on 31/May/23

((n!(n+1)^m )/((n+m)!))=((n!(n+1)^m )/((n+m)(n+m−1)...(n+1)n!)) =((n^m +mn^(m−1) +...+1)/(n^m +(1+2+..+m)n^(m−1) +...+m!)) ⇒lim_(n→∞) ((n!(n+1)^m )/((n+m)!)) =lim_(n→∞) ((n^m +mn^(m−1) +...+1)/(n^m +(1+2+..+m)n^(m−1) +...+m!)) =1 ✓

n!(n+1)m(n+m)!=n!(n+1)m(n+m)(n+m−1)...(n+1)n!=nm+mnm−1+...+1nm+(1+2+..+m)nm−1+...+m!⇒limn→∞n!(n+1)m(n+m)!=limn→∞nm+mnm−1+...+1nm+(1+2+..+m)nm−1+...+m!=1✓

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