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Question Number 131374 by shaker last updated on 04/Feb/21

Answered by MJS_new last updated on 04/Feb/21

n!≈((n/e))^n (√(2πn)) ((((4n)!)/((3n)!)))^(1/n) ≈((((((4n)/e))^(4n) (√(8πn)))/((((3n)/e))^(3n) (√(6πn)))))^(1/n) =(((((4n)/e))^4 )/((((3n)/e))^3 ))((4/3))^(1/n) = =((256n)/(27e))((4/3))^(1/n) ⇒ we have lim_(n→+∞) (ln (((256)/(27e))((4/3))^(1/n) ))= =ln ((256)/(27e)) =8ln 2 −3ln 3 −1

n!≈(ne)n2πn((4n)!(3n)!)1/n≈((4ne)4n8πn(3ne)3n6πn)1/n=(4ne)4(3ne)3(43)1/n==256n27e(43)1/n⇒wehavelimn→+∞(ln(25627e(43)1/n))==ln25627e=8ln2−3ln3−1

Answered by aleks041103 last updated on 04/Feb/21

Stirling′s approximation (n)!∼ n^n e^(−n) =((n/e))^n Proof: ln(x!)=Σ_(y=1) ^x ln(y)=Σ_(y=1) ^x ln(y)Δy≈∫_1 ^x ln(y)dy Δy=1 ∫_1 ^x ln(y)dy=yln(y)∣_1 ^x −∫_1 ^x yd(ln(y))= =xln(x)−∫_1 ^x dy=xln(x)−x+1 x≫1, ln(x!)∼ln(x^x e^(−x) ) Q.E.D lim_(n→∞) [ln((((((4n)!)/((3n)!)))^(1/n) /n))]= =lim_(n→∞) [ln(((((((4^4 n^4 e^3 )/(3^3 n^3 e^4 )))^n ))^(1/n) /n))]= =lim_(n→∞) [ln(((256n)/(27ne)))]=ln(256/27)+ln(1/e)= =ln(((256)/(27)))−1

Stirling′sapproximation(n)!∼nne−n=(ne)nProof:ln(x!)=∑xy=1ln(y)=∑xy=1ln(y)Δy≈∫x1ln(y)dyΔy=1∫x1ln(y)dy=yln(y)∣1x−∫x1yd(ln(y))==xln(x)−∫x1dy=xln(x)−x+1x≫1,ln(x!)∼ln(xxe−x)Q.E.Dlimn→∞[ln((4n)!(3n)!nn)]==limn→∞[ln((44n4e333n3e4)nnn)]==limn→∞[ln(256n27ne)]=ln(256/27)+ln(1/e)==ln(25627)−1

Answered by Dwaipayan Shikari last updated on 04/Feb/21

log((((((4n)^(4n) e^(−4n) (√(8πn)))/((3n)^(3n) e^(−3n) (√(6πn)))))^(1/n) /n))=log(((256)/(27e))((4/3))^(1/n) ) lim_(n→∞) log(((256)/(27e))((4/3))^(1/n) )=log(((256)/(27e)))

log((4n)4ne−4n8πn(3n)3ne−3n6πnnn)=log(25627e(43)1n)limn→∞log(25627e(43)1n)=log(25627e)

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