# `sum_(n=1)^oo 1/5^n` Use the Root Test to determine the convergence or divergence of the series.

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 `

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.

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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`

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.

Approved by eNotes Editorial Team