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

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` since `n! > n^2` for `n>=4`

The limit diverges, so the sum also diverges.

Comments

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