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=2)^oo...
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=2)^oo n/(nln(n))^n` , we have `a_n =n/(nln(n))^n` .
Applying the Root test, we set-up the limit as:
`lim_(n-gtoo) |n/(nln(n))^n|^(1/n) =lim_(n-gtoo) (n/(nln(n))^n)^(1/n)`
Apply Law of Exponent: `(x/y)^n = x^n/y^n` and `(x^n)^m = x^(n*m)` .
`lim_(n-gtoo) (n/(nln(n))^n)^(1/n) =lim_(n-gtoo) n^(1/n)/((nln(n))^n)^(1/n)`
`=lim_(n-gtoo) n^(1/n)/(nln(n))^(n*1/n)`
`=lim_(n-gtoo) n^(1/n)/(nln(n))^(n/n)`
`=lim_(n-gtoo) n^(1/n)/(nln(n))^1`
`=lim_(n-gtoo) n^(1/n)/(nln(n))`
Apply the limit property: `lim_(x-gta)[(f(x))/(g(x))] =(lim_(x-gta) f(x))/(lim_(x-gta) g(x)).`
`lim_(n-gtoo) n^(1/n)/(nln(n))=(lim_(n-gtoo) n^(1/n))/(lim_(n-gtoo) nln(n))`
`=1/ oo`
`= 0 `
Note: `lim_(n-gtoo) n^(1/n) = 1 ` and
` lim_(n-gtoo) nln(n) = oo ln(oo)`
` = oo*oo`
` =oo`
The limit value `L=0` satisfies the condition: `L lt1` since `0lt1` .
Conclusion: The series `sum_(n=2)^oo n/(nln(n))^n` is absolutely convergent.
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