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=0)^oo 5^n/(2^n+1)` , we let:

`a_n =5^n/(2^n+1)`

We set-up the limit as:

`lim_(n-gtoo) |5^n/(2^n+1)|^(1/n)=lim_(n-gtoo) (5^n/(2^n+1))^(1/n)`

Apply Law of Exponent: `(x/y)^n = x^n/y^n` and `(x^n)^m = x^(n*m)` .

`lim_(n-gtoo) (5^n/(2^n+1))^(1/n)=lim_(n-gtoo) (5^n)^(1/n)/(2^n+1)^(1/n)`

`=lim_(n-gtoo) 5^(n/n)/(2^n+1)^(1/n)`

`=lim_(n-gtoo) 5^1/(2^n+1)^(1/n)`

`=lim_(n-gtoo) 5/(2^n+1)^(1/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) 5/(2^n+1)^(1/n) =(lim_(n-gtoo) 5)/(lim_(n-gtoo) (2^n+1)^(1/n))`

` = 5 / 2`

Note: Applying `a^x =e^(xln(a))` , we may let: `(2^n+1)^(1/n) = e^(1/nln(2^n+1))`

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

Apply L'Hospital's rule:

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

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

`=oo/oo`

Apply again the L'Hospital's rule:

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

`=lim_(n-gtoo) (ln(2))`

` = ln(2)`

Applying `lim_(n-gtoo)1/nln(2^n+1)= ln(2) on e^(1/nln(2^n+1))` , we get:

`lim_(n-gtoo) e^(1/nln(2^n+1)) = e^(ln(2)) = 2`

The limit value `L = 5/2 or 2.5 ` satisfies the condition: `Lgt1 ` since `2.5gt1` .

Conclusion: The series `sum_(n=0)^oo 5^n/(2^n+1)` is **divergent**.