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=nn√n!an+1≥an⟺
n+1n+1√(n+1)!≥nn√n!⟺n+1(n+1)!(1n+1)n!(1n)≥n⟺
(n+1)(1−1n+1)n!(1n−1n+1)≥n⟺(n+1)(nn+1)n!(1n(n+1))≥n
(n+1)n2n!≥n(n(n+1))=nn2nn⟺nnn!≤(n+1n)n2=(1+1n)n2⟺
(1+1n)n2≥nnn!
which is true since
(1+1n)n2≥(1+1n)n2−1=[(1+1n)n+1]n−1≥en−1≥nnn!
indeed
en−1≥nnn!⟺bn=enn!nn≥e
which is true ∀n since
n=1⟹b1=e11!11≥e
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
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