limn→∞(1+1n!)2n
I've tried:
limn→∞(1+1n!)2n=limn→∞eln(1+1n!)2n=limn→∞e2n ln(1+1n!)
But I don't know how to work with the factorial
Answer
limn→∞(1+1n!)2n=limn→∞((1+1n!)n!)2(n−1)!=e0=1
Alternatively,
Let A=(1+1n!)2n
So, logA=2nlog(1+1n!) using log(1+x)=x−x22+x33−x44+⋯
which comes from the Taylor series which holds for |x|<1 as n→∞,1n!→0⟹|limn→∞1n!|<1
=2n(1n!−1(n!)22+1(n!)33+⋯)
=2(n−1)!−1n!(n−1)!+23(n−1)!(n!)2+⋯
So, limn→∞logA=0⟹limn→∞A=e0=1
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