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.
For the given...
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.
For the given series `sum_(n=1)^oo (ln(n)/n)^n` , we have `a_n =(ln(n)/n)^n` .
Applying the Root test, we set-up the limit as:
`lim_(n-gtoo) |(ln(n)/n)^n|^(1/n) =lim_(n-gtoo) ((ln(n)/n)^n)^(1/n)`
Apply Law of Exponent: `(x^n)^m = x^(n*m)` .
`lim_(n-gtoo) ((ln(n)/n)^n)^(1/n)=lim_(n-gtoo) (ln(n)/n)^(n*1/n)`
`=lim_(n-gtoo) (ln(n)/n)^(n/n)`
`=lim_(n-gtoo) (ln(n)/n)^1`
`=lim_(n-gtoo) (ln(n)/n)`
Evaluate the limit using direct substitution: `n = oo` .
`lim_(n-gtoo) (ln(n)/n) = oo/oo `
When the limit value is indeterminate `(oo/oo)` , we may apply L'Hospital's Rule:
`lim_(x-gta) (f(x))/(g(x)) =lim_(x-gta) (f'(x))/(g'(x))` .
Let: `f(n) = ln(n)` then
`g(n) = n` then `g'(n) =1` .
Then, the limit becomes:
`lim_(n-gtoo) (ln(n)/n)=lim_(n-gtoo) ((1/n))/1`
` =lim_(n-gtoo) 1/n`
` = 1/oo`
` =0`
The limit value `L=0` satisfies the condition: `L lt1` since `0lt1.`
Therefore, the series sum_(n=1)^oo (ln(n)/n)^n is absolutely convergent.
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