Calculus of a Single Variable, Chapter 9, 9.6, Section 9.6, Problem 24

In using 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.
We may apply the Root Test to determine the convergence or divergence of the series sum_(n=1)^oo (-1)^(n-1) *(3/2)^n/n^2 .
lim_(n-gtoo) |((-1)^(n-1) *(3/2)^n/n^2 )^(1/n)| =lim_(n-gtoo) |(-1)^((n-1)*1/n)(3/2)^(n*1/n)/n^(2*1/n)|
=lim_(n-gtoo) |(-1)^(n/n-1/n)(3/2)^(n/n)/n^(2/n)|
=lim_(n-gtoo)( 1 * (3/2)^1/n^(2/n))
=lim_(n-gtoo) (3/2)/n^(2/n)
Note: |(-1)^(n/n-1/n)| = 1
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)/n^(2/n)=(lim_(n-gtoo) 3/2)/(lim_(n-gtoo)n^(2/n))
= ((3/2))/1
=3/2 or 1.5
The limit value L = 3/2 or 1.5 satisfies the condition: Lgt1 since 3/2gt 1 or 1.5gt1 .
Thus, the series sum_(n=1)^oo (-1)^(n-1) *(3/2)^n/n^2 is divergent.