How do you determine whether the sequence #a_n=(n!)^(1/n)# converges, if so how do you find the limit?

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Answer 1 · 2017-10-05T22:25:19

#lim_(n->oo) (n!)^(1/n) = oo#

Consider:

#(n!)^2 = (n(n-1)(n-2) * ...* 2*1) xx (n(n-1)(n-2)*...*2*1)#

Rearranging the factors:

#(n!)^2 = (n * 1) * ( (n-1) * 2 ) * ... * (2 * (n-2)) * (n * 1)#

#(n!)^2 = prod_(k=0)^(n-1) (n-k)(k+1)#

For each factor:

#(n-k)(k+1) = kn+n-k^2-k#

#(n-k)(k+1) = n+k(n-k-1) >= n#

So:

#(n!)^2 >= n^n#

and therefore:

#(n!)^(1/n) >= sqrt n#

Then:

#lim_(n->oo) (n!)^(1/n) = oo#