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_(n=1)^oo ((2n)/(n+1))^n` , we
let:

`a_n =((n-2)/(5n+1))^n`

We set-up the limit as:

`lim_(n-gtoo) |((n-2)/(5n+1))^n|^(1/n) =lim_(n-gtoo) (((n-2)/(5n+1))^n)^(1/n)`

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

`lim_(n-gtoo) (((n-2)/(5n+1))^n)^(1/n) =lim_(n-gtoo) ((n-2)/(5n+1))^(n*(1/n) )`

`=lim_(n-gtoo) ((n-2)/(5n+1))^(n/n )`

`=lim_(n-gtoo) ((n-2)/(5n+1))^1`

`=lim_(n-gtoo) (n-2)/(5n+1)`

Evaluate the limit.

`lim_(n-gtoo)(n-2)/(5n+1) =lim_(n-gtoo)(n/n-2/n)/((5n)/n+1/n)`

` =lim_(n-gtoo)(1-2/n)/(5+1/n)`

` =(lim_(n-gtoo)(1-2/n))/(lim_(n-gtoo)(5+1/n))`

` = (1-2/oo)/(5+1/oo)`

` =(1-0)/(5+0)`

` =1/5 or 0.2"`

The limit value `L =1/5` `or 0.2` satisfies the condition: `Llt1` since `1/5 lt1` or `0.2 lt1` .

Thus, the series `sum_(n=1)^oo ((2n)/(n+1))^n` is **absolutely
convergent**.

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