`sum_(n=1)^oo (-1)^(n-1)(3/2)^n/n^2` Use the Root Test to determine the convergence or divergence of the series.

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marizi eNotes educator| Certified Educator

In using Root test on a series sum a_n, we determine the limit as:

`lim_(n-gtoo) root(n)(|a_n|)= L`


`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