How do you use the Root Test on the series $\infty \sum n = 1 {(\frac{n!}{n})}^{n}$ $sum_(n=1)^oo((n!)/n)^n$ ?

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Answer 1 · 2014-09-22T01:08:03

By Root Test,

$\infty \sum n = 1 {(\frac{n!}{n})}^{n}$ $sum_{n=1}^infty({n!}/{n})^n$ diverges.

Let us look at some details.

Let $a_{n} = {(\frac{n!}{n})}^{n} = {[(n - 1)!]}^{n}$ $a_n=({n!}/n)^n=[(n-1)!]^n$ .

Consider:

$lim_{n to infty}rootn{|a_n|}=lim_{n to infty}rootn{[(n-1)!]^n} =lim_{n to infty}(n-1)! =infty$ ,

which is greater than 1; therefore, we can conclude that the series diverges by Root Test.

How do you use the Root Test on the series sum_(n=1)^oo((n!)/n)^n∞∑n=1(n!n)nsum_(n=1)^oo((n!)/n)^n ?