Thursday, December 5, 2019

real analysis - Monotonicity of fracnsqrt[n](n!)



It is known that when n the sequence nn(n!) has limit e but I don't know how to prove it's monotonicity. After a short calculus using WolframAlpha I found that this sequence is actually increasing. I tried to compare 2 consecutive members but I couldn't manage to show something. It is obvious that nnn! is increasing but that don't help us much (I think)



Answer



We need to show that



an=nnn!an+1an



n+1n+1(n+1)!nnn!n+1(n+1)!(1n+1)n!(1n)n



(n+1)(11n+1)n!(1n1n+1)n(n+1)(nn+1)n!(1n(n+1))n



(n+1)n2n!n(n(n+1))=nn2nnnnn!(n+1n)n2=(1+1n)n2



(1+1n)n2nnn!



which is true since




(1+1n)n2(1+1n)n21=[(1+1n)n+1]n1en1nnn!



indeed



en1nnn!bn=enn!nne



which is true n since



n=1b1=e11!11e




and



bn+1bn=en+1(n+1)!(n+1)n+1nnenn!=e(1+1n)n>1



For the last inequality see also the related OP



Show that e1nn!nn


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