`sum_(n=1)^oo (-1)^n/sqrt(n)` Determine the convergence or divergence of the series.

To determine the convergence or divergence of the series `sum_(n=1)^oo (-1)^n/sqrt(n)` , we may apply the Root Test.

In Root test, 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) `L lt1 ` then the series converges absolutely.

b) `Lgt1 ` then the series diverges.

c) `L=1` or does not exist  then the test is inconclusive.The series may be divergent, conditionally convergent, or...

To determine the convergence or divergence of the series `sum_(n=1)^oo (-1)^n/sqrt(n)` , we may apply the Root Test.

In Root test, 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) `L lt1 ` then the series converges absolutely.

b) `Lgt1 ` then the series diverges.

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 (-1)^n/sqrt(n)` , we have `a_n =(-1)^n/sqrt(n)` .

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

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

Note: `|(-1)^n| = 1`

Apply radical property: `root(n)(x) =x^(1/n)` and Law of exponent: `(x/y)^n = x^n/y^n.`

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

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

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

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

Apply the limit property:` lim_(x-gta)[(f(x))/(g(x))] =(lim_(x-gta) f(x))/(lim_(x-gta) g(x)).`

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

                     ` = 1/1`

                     ` =1`

The limit value `L = 1` implies that the series may be divergent, conditionally convergent, or absolutely convergent.

To verify, we use alternating series test on `sum (-1)^n a_n` .

`a_n = 1/sqrt(n) ` is positive and decreasing from `N=1` .

`lim_(n-gtoo)1/sqrt(n) = 1/oo = 0`

Based on alternating series test condition,  the series `sum_(n=1)^oo (-1)^n/sqrt(n)` converges.

Apply p-series test on `sum |a_n|` .

`sum_(n=1)^oo |(-1)^n/sqrt(n)|=sum_(n=1)^oo 1/sqrt(n)`

                     ` =sum_(n=1)^oo 1/n^(1/2)`

Based on p-series test condition,  we have `p=1/2` that satisfies `0ltplt=1` .

Thus, the series  `sum_(n=1)^oo |(-1)^n/sqrt(n)| ` diverges.      

Notes:

In p-series test, we follow:

-if `p` is within the interval of `0ltplt=1 ` then the series diverges.

-if `p ` is within the interval of `pgt1 ` then the series converges.

Absolute Convergence:  `sum a_n`  is absolutely convergent if `sum|a_n|`  ` `is convergent.             

Conditional Convergence:  `sum a_n` is conditionally convergent if `sum|a_n|` ` ` is divergent and `sum a_n` ` `is convergent.       

Conclusion:

The series `sum_(n=1)^oo (-1)^n/sqrt(n)` is conditionally convergent since  `sum_(n=1)^oo (-1)^n/sqrt(n)` is convergent  and `sum_(n=1)^oo |(-1)^n/sqrt(n)|` is divergent.