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Question-192927

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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
$$\mathrm{n}!\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{m}} \approx\sqrt{\mathrm{2}\pi\mathrm{n}}\left(\frac{\mathrm{n}}{\mathrm{e}}\right)^{\mathrm{n}} ×\mathrm{n}^{\mathrm{m}} \:\:\wedge\:\left(\mathrm{n}+\mathrm{m}\right)!\approx\sqrt{\mathrm{2}\pi\left(\mathrm{n}+\mathrm{m}\right)}\left(\frac{\mathrm{n}+\mathrm{m}}{\mathrm{e}}\right)^{\mathrm{n}+\mathrm{m}} \\ $$$$\Rightarrow\frac{\mathrm{n}!\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{m}} }{\left(\mathrm{n}+\mathrm{m}\right)!}\approx\frac{\sqrt{\mathrm{2}\pi\mathrm{n}}}{\:\sqrt{\mathrm{2}\pi\left(\mathrm{n}+\mathrm{m}\right)}}×\frac{\mathrm{n}^{\mathrm{n}+\mathrm{m}} }{\mathrm{e}^{\mathrm{n}} }×\frac{\mathrm{e}^{\mathrm{n}+\mathrm{m}} }{\left(\mathrm{n}+\mathrm{m}\right)^{\mathrm{n}+\mathrm{m}\:} } \\ $$$$\Rightarrow\frac{\mathrm{n}!\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{m}} }{\left(\mathrm{n}+\mathrm{m}\right)!}\approx\frac{\sqrt{\mathrm{n}}}{\:\sqrt{\mathrm{n}+\mathrm{m}}}×\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{m}}\right)^{\mathrm{n}+\mathrm{m}} ×\mathrm{e}^{\mathrm{m}} \\ $$$$\Rightarrow\frac{\mathrm{n}!\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{m}} }{\left(\mathrm{n}+\mathrm{m}\right)!}\approx\sqrt{\mathrm{1}−\frac{\mathrm{m}}{\mathrm{n}+\mathrm{m}}}×\left(\mathrm{1}−\frac{\mathrm{m}}{\mathrm{n}+\mathrm{m}}\right)^{\mathrm{n}+\mathrm{m}} ×\mathrm{e}^{\mathrm{m}} \\ $$$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{n}!\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{m}} }{\left(\mathrm{n}+\mathrm{m}\right)!}=\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\sqrt{\mathrm{1}−\frac{\mathrm{m}}{\mathrm{n}+\mathrm{m}}}×\left(\mathrm{1}−\frac{\mathrm{m}}{\mathrm{n}+\mathrm{m}}\right)^{\mathrm{n}+\mathrm{m}} ×\mathrm{e}^{\mathrm{m}} =\mathrm{1}×\mathrm{e}^{−\mathrm{m}} ×\mathrm{e}^{\mathrm{m}} =\mathrm{1} \\ $$$$ \\ $$
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 ✓
$$\frac{{n}!\left({n}+\mathrm{1}\right)^{{m}} }{\left({n}+{m}\right)!}=\frac{{n}!\left({n}+\mathrm{1}\right)^{{m}} }{\left({n}+{m}\right)\left({n}+{m}−\mathrm{1}\right)…\left({n}+\mathrm{1}\right){n}!} \\ $$$$=\frac{{n}^{{m}} +{mn}^{{m}−\mathrm{1}} +…+\mathrm{1}}{{n}^{{m}} +\left(\mathrm{1}+\mathrm{2}+..+{m}\right){n}^{{m}−\mathrm{1}} +…+{m}!}\: \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} \:\frac{{n}!\left({n}+\mathrm{1}\right)^{{m}} }{\left({n}+{m}\right)!} \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{{n}^{{m}} +{mn}^{{m}−\mathrm{1}} +…+\mathrm{1}}{{n}^{{m}} +\left(\mathrm{1}+\mathrm{2}+..+{m}\right){n}^{{m}−\mathrm{1}} +…+{m}!}\:=\mathrm{1}\:\checkmark \\ $$$$ \\ $$$$ \\ $$

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