Saturday, February 3, 2018

calculus - Help solving the limit limntoinftyleft(1+frac1n!right)2n



limn(1+1n!)2n



I've tried:

limn(1+1n!)2n=limneln(1+1n!)2n=limne2n ln(1+1n!)


But I don't know how to work with the factorial


Answer



limn(1+1n!)2n=limn((1+1n!)n!)2(n1)!=e0=1






Alternatively,
Let A=(1+1n!)2n




So, logA=2nlog(1+1n!) using log(1+x)=xx22+x33x44+



which comes from the Taylor series which holds for |x|<1 as n,1n!0|limn1n!|<1



=2n(1n!1(n!)22+1(n!)33+)



=2(n1)!1n!(n1)!+23(n1)!(n!)2+



So, limnlogA=0limnA=e0=1


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