Calculus of a Single Variable, Chapter 9, 9.6, Section 9.6, Problem 44

To apply the 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.
To apply the Root Test to determine the convergence or divergence of the series sum_(n=0)^oo e^(-3n) , we let a_n = e^(-3n) .
Apply Law of Exponent: x^(-n) = 1/x^n .
a_ n = 1/e^(3n).
Applying the Root test, we set-up the limit as:
lim_(n-gtoo) |1/e^(3n)|^(1/n) =lim_(n-gtoo) (1/e^(3n))^(1/n)
Apply the Law of Exponents: (x/y)^n = x^n/y^n and (x^n)^m= x^(n*m) .
lim_(n-gtoo) (1/e^(3n))^(1/n) =lim_(n-gtoo) 1^(1/n)/(e^(3n))^(1/n)
=lim_(n-gtoo) 1^(1/n)/e^(3n*1/n)
=lim_(n-gtoo) 1^(1/n)/e^((3n)/n)
=lim_(n-gtoo) 1^(1/n)/e^3
Apply the limit property: lim_(x-gta)[(f(x))/(g(x))] =(lim_(x-gta) f(x))/(lim_(x-gta) g(x)) .
lim_(n-gtoo) 1^(1/n)/e^3 =(lim_(n-gtoo) 1^(1/n))/(lim_(n-gtoo)e^3 )
= 1^(1/oo) /e^3
=1^0/e^3
=1/e^3 or 0.0498 (approximated value)
The limit value L = 1/e^3 or 0.0498 satisfies the condition: Llt1 since 0.0498lt1.
Thus, the series sum_(n=0)^oo e^(-3n) is absolutely convergent.