To apply **Root test** on a series `sum a_n` , we
determine the limit as:

`lim_(n-gtoo) root(n)(|a_n|)= L`

or

`lim_(n-gtoo) |a_n|^(1/n)= L`

Then, we follow the conditions:

a) `Llt1` then the series is **absolutely
convergent**.

b) `Lgt1 ` then the series is **divergent**.

c) `L=1 ` or *does not exist* then
the **test is inconclusive**. The series may be divergent,
conditionally convergent, or absolutely convergent.

In order to apply **Root Test** in determining the
convergence or divergence of
the **series** `sum_(n=1)^oo ((3n+2)/(n+3))^n` , we let:
`a_n =((3n+2)/(n+3))^n.`

We set-up the limit as:

`lim_(n-gtoo) |((3n+2)/(n+3))^n|^(1/n) =lim_(n-gtoo) (((3n+2)/(n+3))^n)^(1/n) `

Apply the Law of Exponents: `(x^n)^m= x^(n*m)` .

`lim_(n-gtoo) (((3n+2)/(n+3))^n)^(1/n) =lim_(n-gtoo) ((3n+2)/(n+3))^(n*1/n)`

`=lim_(n-gtoo) ((3n+2)/(n+3))^(n/n)`

`=lim_(n-gtoo) ((3n+2)/(n+3))^1`

`=lim_(n-gtoo) (3n+2)/(n+3)`

To evaluate the limit `lim_(n-gtoo) (3n+2)/(n+3)` , we divide each term by the highest denominator power: `n` .

`lim_(n-gtoo) (3n+2)/(n+3)=lim_(n-gtoo)((3n)/n+2/n)/(n/n+3/n)`

`=lim_(n-gtoo) (3+2/n)/(1+3/n)`

Apply the limit property: `lim_(x-gta)[(f(x))/(g(x))] =(lim_(x-gta) f(x))/(lim_(x-gta) g(x)). `

`lim_(n-gtoo) (3+2/n)/(1+3/n) =(lim_(n-gtoo) (3+2/n))/(lim_(n-gtoo)(1+3/n)) `

` = (3+2/oo)/(1+3/oo)`

` = (3+0)/(1+0)`

` =3/1 `

` =3`

The limit value `L = 3 ` satisfies the condition: `Lgt1` since `3gt1` .

Conclusion: The series `sum_(n=1)^oo ((3n+2)/(n+3))^n` is
**divergent.**

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