`sum_(n=0)^oo (-1)^n*e^(-n^2)` Determine whether the series converges absolutely or conditionally, or diverges.

Expert Answers
marizi eNotes educator| Certified Educator

To apply 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.

In order to apply Root Test in determining the convergence or divergence of the series =`sum_(n0)^oo (-1)^n*e^(-n^2)` , we let:


We set-up the limit as: 

`lim_(n-gtoo) |(-1)^n*e^(-n^2)|^(1/n) =lim_(n-gtoo) |(-1)^n|^(1/n)*|e^(-n^2)|^(1/n)`

                                         ` =lim_(n-gtoo) 1 *(e^(-n^2))^(1/n) `

                                         ` =lim_(n-gtoo) (e^(-n^2))^(1/n)`

Apply the Law of Exponents:  `(x^n)^m= x^(n*m)` and `x^(-n)= 1/x^n` .

`lim_(n-gtoo) (e^(-n^2))^(1/n) =lim_(n-gtoo)e^(-n^2*1/n)`



                              ` =lim_(n-gtoo)1/e^n`

Evaluate the limit.

`lim_(n-gtoo)1/e^n = 1/e^oo`

                  ` = 1/oo`

                  ` =0`

The limit value `L=0` satisfies the condition: `L lt1` since `0lt1` .

 Therefore, the series `sum_(n=0)^oo(-1)^n*e^(-n^2) ` is absolutely convergent.