# `sum_(n=0)^oo 5^n/(2^n+1)` Use the Root Test to determine the convergence or divergence of the series.

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`

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.

Approved by eNotes Editorial Team