Find limn→∞(n!)1/nn.
I don't know how to start. Hints are also appreciated.
Answer
let y=(n!)1/nn.
⟹y=(n(n−1)(n−2)....3.2.1n.n.n....nnn)1n
Note that we can distribute the n in the denominator and give an ′n′ to each term
⟹logy=1n(log1n+log2n+...+lognn)
applying limn→∞ on both sides, we find that R.H.S is of the form
limn→∞1nr=n∑r=0f(rn)
which can be evaluated by integration =∫10log(x)dx
=xlogx−x
plugging in the limits (carefully here) we get −1.
logy=−1,⟹y=1e
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