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What is `lim_(n->oo)(((n!)^2)/(n^(2n)))^(1/n)`

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hhubbes | (Level 1) Honors

Posted February 16, 2012 at 5:56 AM via web

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What is `lim_(n->oo)(((n!)^2)/(n^(2n)))^(1/n)`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 16, 2012 at 12:02 PM (Answer #1)

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The limit `lim_(n->oo)(((n!)^2)/(n^(2n)))^(1/n)` has to be determined.

` ` ` `n! = 1*2*3*...n < n^n which is n multiplied by itself n times.

As the denominator is greater than the numerator the term `((n!)^2)/(n^(2n))` decreases as n increases.

As n tends to inf., `((n!)^2)/(n^(2n))` tends to 0.Also, `(1/n)` tends to 0

The required limit `lim_(n->oo)(((n!)^2)/(n^(2n)))^(1/n)`

reduces to the form

`lim_(n->0) n^n` = 1

The limit `lim_(n->oo)(((n!)^2)/(n^(2n)))^(1/n)` = 1

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