Calculus of a Single Variable, Chapter 9, 9.6, Section 9.6, Problem 42

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)/(2n+1))^(3n) , we let:
a_n =((-3n)/(2n+1))^(3n)
We set-up the limit as:
lim_(n-gtoo) |((-3n)/(2n+1))^(3n)|^(1/n) =lim_(n-gtoo) (((3n)/(2n+1))^(3n))^(1/n) .
Note: | - f(n)| = f(n) .
Apply the Law of Exponents: (x^n)^m= x^(n*m).
lim_(n-gtoo) (((3n)/(2n+1))^(3n))^(1/n)=lim_(n-gtoo) ((3n)/(2n+1))^(3n*(1/n))
=lim_(n-gtoo) ((3n)/(2n+1))^((3n)/n)
=lim_(n-gtoo) ((3n)/(2n+1))^3
Apply the limit property: lim_(x-gta)[f(x)]^b =[lim_(x-gta)f(x)]^b such that b is a real constant.
lim_(n-gtoo) ((3n)/(2n+1))^3 =(lim_(n-gtoo) ((3n)/(2n+1)))^3
To evaluate the limit lim_(n-gtoo) (3n)/(2n+1) , we divide each term by the highest denominator power: n.
lim_(n-gtoo) (3n)/(2n+1) =lim_(n-gtoo) ((3n)/n)/((2n)/n+1/n)
=lim_(n-gtoo) 3/(2+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) 3/(2+1/n) =(lim_(n-gtoo) 3)/(lim_(n-gtoo) (2+1/n))
=3/(2+1/oo)
=3/(2+0)
=3/2
Applying lim_(n-gtoo) (3n)/(2n+1)=3/2 , we get:
(lim_(n-gtoo) ((3n)/(2n+1)))^3 =(3/2)^3
Apply Law of Exponent: (x/y)^n = x^n/y^n and simplify.
(3/2)^3 =3^3/2^3
= 27/8 or 3.375
The limit value L = 27/8 or 3.375 satisfies the condition: Lgt1 .
Thus, the series sum_(n=1)^oo ((-3n)/(2n+1))^(3n) is divergent.