lim n->inf [n^(3n) / (n!)^3 ]^(1/n)

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beckden | High School Teacher | (Level 1) Educator

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`lim_(n->oo)(n^(3n)/(n!)^3)^(1/n)`

`=lim_(n->oo)(n^n/(n!))^(3/n)`

We know that `````int^n_1 ln(x) dx <= sum_(j=1)^n ln(j) lt= int^n_0 ln(x+1) dx`

``so `n ln(n/e) + 1 <= ln n! <= (n+1) ln((n+1)/e) + 1` and this reduces to

`e(n/e)^n lt= n! lt= e((n+1)/e)^(n+1)`

`en^n/e^n lt= n! lt= e(n+1)^(n+1)/e^n`

So since all of these values are positive we can take the reciprocal and get

`e^(n-1)/(n+1)^(n+1) lt= 1/(n!) lt= e^(n-1)/n^n`

`(e^(n-1)n^n)/(n+1)^(n+1) lt= n^n/(n!) lt= e^(n-1)`

taking the n'th root we get

`(e^((n-1)/n)n)/((n+1)^((n+1)/n)) <= (n^n/(n!))^(1/n) <= e^((n-1)/n)`

The limit of both sides as `n->oo` is `e` so by the sandwich theorem shows that

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

So we get

`lim_(n->oo) (n^(3n))/(n!)^3 = e^3`

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