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_(n0)^oo (-1)^n*e^(-n^2)` , we let:

`a_n=(-1)^n*e^(-n^2)`

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)e^(-n^2/n)`

`=lim_(n-gtoo)e^(-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.**

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