Calculus of a Single Variable, Chapter 9, 9.6, Section 9.6, Problem 31

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.