Wednesday, 11 May 2016

`sum_(n=1)^oo (lnn/n)^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`


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.

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

While class does play a role in Rich and Humble , race and class aren't addressed by William Taylor Adams (Oliver Opic's real name) ...