# `sum_(n=1)^oo 5^n/n^4` 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)...

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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 Root test on the given series `sum_(n=1)^oo 5^n/n^4` when we let:  `a_n =5^n/n^4` .

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

`lim_(n-gtoo) |5^n/n^4|^(1/n) =lim_(n-gtoo) (5^n/n^4)^(1/n)`

Apply Law of Exponent: `(x/y)^n = x^n/y^n` and `(x^n)^m= x^(n*m)` .

`lim_(n-gtoo) (5^n/n^4)^(1/n) =lim_(n-gtoo) (5^n)^(1/n)/(n^4)^(1/n)`

` =lim_(n-gtoo)5^(n*1/n)/n^(4*1/n)`

` =lim_(n-gtoo)5^(n/n)/n^(4/n)`

` =lim_(n-gtoo)5^1/n^(4/n)`

` =lim_(n-gtoo)5/n^(4/n)`

Evaluate the limit.

`lim_(n-gtoo) 5/n^(4/n)=5 lim_(n-gtoo) 1/n^(4/n) `

` =5 *1/oo^(4/oo)`

` =5 *1/oo^(0)`

` =5 *1/1`

` = 5*1`

` =5`

The limit value `L =5` satisfies the condition: `Lgt1` since `5gt1` .

Conclusion: The series `sum_(n=1)^oo 5^n/n^4` is divergent.

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