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`sum_(n=1)^oo (n!)^n/(n^n)^2` Use the Root Test to determine the convergence or divergence of the series.

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The Root Test says that this limit will converge if and only if the following limit (actually limit superior) is less than 1:
`C = lim_{n rightarrow infty} |(n!)^n/(n^n)^2|^(1/n)`

This is actually a fairly simple limit to get:

`C = lim_(n->oo) |{n!}/n^2| = infty` sinceĀ `n! > n^2` for `n>=4`


The limit diverges, so the sum also diverges.

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