# `sum_(n=1)^oo (-1)^n/sqrt(n)` Determine whether the series converges absolutely or conditionally, or diverges.

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`

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 a_n` .

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

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

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.

Conclusion:

`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.