Question

How do you apply the ratio test to determine if `Sigma (3^n(n!))/(n^n)` from `n=[1,oo)` is convergent to divergent?

Answer 1

It is divergent.

Explanation:

We can apply d'Alembert's ratio test:

Suppose that;

`S=sum_(r=1)^oo a_n \ \`, and `\ \ L=lim_(n rarr oo) |a_(n+1)/a_n|`

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

`S=sum_(n=1)^oo (3^n(n!))/n^n`

So our test limit is:

`L = lim_(n rarr oo) | {(3^(n+1)((n+1)!))/(n+1)^(n+1)}/{(3^n(n!))/n^n}|`
`\ \ \ = lim_(n rarr oo) | (3^(n+1)(n+1)!)/(n+1)^(n+1) * n^n/(3^n n!}|`
`\ \ \ = lim_(n rarr oo) | (3 * 3^n(n+1) * n!)/(n+1)^(n+1) * n^n/(3^n n!}|`
`\ \ \ = lim_(n rarr oo) | (3 (n+1))/(n+1)^(n+1) * n^n|`
`\ \ \ = lim_(n rarr oo) | (3n^n )/(n+1)^n |`
`\ \ \ = 3 \ lim_(n rarr oo) | (n/(n+1))^n |`
`\ \ \ = 3 \ lim_(n rarr oo) (n/(n+1))^n`
`\ \ \ = 3/e`
`\ \ \ ~~ 1.1 > 1`

Hence the series is divergent.

In case you are wondering how the final limit was established; it is a standard limit in disguise:

Let `u=n+1 => n=u-1` then `u rarr oo` as `n rarr oo`, And

`lim_(n rarr oo) (n/(n+1))^n = lim_(u rarr oo) ((u-1)/u)^(u-1)`
`" " = lim_(u rarr oo) (1-1/u)^u / (1-1/u)`
`" " = {lim_(u rarr oo) (1-1/u)^u }/ {lim_(u rarr oo)(1-1/u)}`
`" " = lim_(u rarr oo) (1-1/u)^u`

Which courtesy of Leonhard Euler is standard limit with known value `1/e`.

Answer 2

See below.

Explanation:

Using Strirling approximation

`log n! =nlogn-n+O(logn)`

we have

`(3^n(n!))/(n^n) approx (3/e)^n` but as `3/e > 1` the series diverges.