To determine the convergence or divergence of a series `sum a_n` using **Root test**, we evaluate a 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 1/5^n.`

For the given series `sum_(n=1)^oo 1/5^n,` we have `a_n =1/5^n.`

Applying the Root test, we set-up the limit as:

`lim_(n-gtoo) |1/5^n|^(1/n)=lim_(n-gtoo) (1/5^n)^(1/n)`

Apply the Law of Exponents: `(x/y)^n =(x^n/y^n)` .

`lim_(n-gtoo) (1/5^n)^(1/n)=lim_(n-gtoo) 1^(1/n)/5^(n*(1/n))`

` =lim_(n-gtoo) 1^(1/n)/5^(n/n)`

` =lim_(n-gtoo) 1^(1/n)/5^1`

` =lim_(n-gtoo) 1^(1/n)/5`

Evaluate the limit.

`lim_(n-gtoo) 1^(1/n)/5 =1/5 lim_(n-gtoo) 1^(1/n)`

` = 1/5*1^(1/oo)`

` = 1/5 * 1^0`

` = 1/5*1`

` =1/5`

The limit value `L =1/5` satisfies the condition: `Llt1` .

Thus, the series `sum_(n=1)^oo 1/5^n` is **absolutely convergent**.

To apply the **Root test** , 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 1/5^n` .

For the given series `sum_(n=1)^oo 1/5^n` , we have `a_n =1/5^n` .

Applying the Root test, we set-up the limit as:

`lim_(n-gtoo) |1/5^n|^(1/n) =lim_(n-gtoo) (1/5^n)^(1/n)`

Apply the Law of Exponents: `(x/y)^n =(x^n/y^n)` .

`lim_(n-gtoo) (1/5^n)^(1/n)=lim_(n-gtoo) 1^(1/n)/5^(n*(1/n))`

` =lim_(n-gtoo) 1^(1/n)/5^(n/n)`

` =lim_(n-gtoo) 1^(1/n)/5^1`

` =lim_(n-gtoo) 1^(1/n)/5`

Evaluate the limit.

`lim_(n-gtoo) 1^(1/n)/5 =1/5 lim_(n-gtoo) 1^(1/n)`

` = 1/5*1^(1/oo)`

` = 1/5 * 1^0`

` = 1/5*1`

` =1/5`

The limit value `L =1/5` satisfies the condition: `Llt1` .

Conclusion: The series `sum_(n=1)^oo 1/5^n` is **absolutely convergent**.

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