To determine the convergence or divergence of the series `sum_(n=1)^oo
(-1)^(n+1)/(nsqrt(n))` , we may apply **Alternating Series
Test**.

In Alternating Series Test, the series `sum (-1)^(n+1) a_n ` is convergent if:

1) `a_ngt=0`

2) ` a_n` is monotone and decreasing sequence.

3) `lim_(n-gtoo) a_n =0`

For the series `sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n))` , we have:

`a_n = 1/(nsqrt(n))`

Apply the radical property: `sqrt(x) =x^(1/2)` and Law of Exponents: `x^n*x^m =x^(n+m).`

`a_n = 1/(nsqrt(n))`

`=1/(n*n^(1/2))`

`=1/n^(1+1/2)`

`=1/n^(3/2)`

The `a_n =1/n^(3/2) ` is a decreasing sequence.

Then, we set-up the limit as :

`lim_(n-gtoo)1/n^(3/2) = 1/oo =0`

By alternating series test criteria, the series `sum_(n=1)^oo
(-1)^(n+1)/(nsqrt(n))` **converges**.

The series `sum_(n=1)^oo (-1)^(n+1)/(nsqrt(n))` has **positive
and negative elements**. Thus, we must verify if the series converges
absolutely or conditionally. Recall:

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

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

We evaluate the `sum |a_n` | as :

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

`=sum_(n=1)^oo 1/n^(3/2)`

Apply the **p-series test** `sum_(n=1)^oo 1/n^p` is convergent
if `pgt1` and divergent if `0ltplt=1` .

The series `sum_(n=1)^oo 1/n^(3/2)` has `p=3/2 or 1.5` which satisfies `pgt1` . Thus, the `sum_(n=1)^oo |(-1)^(n+1)/(nsqrt(n))|` is convergent.

**Conclusion:**

Based on Absolute convergence criteria, the series `sum_(n=1)^oo
(-1)^(n+1)/(nsqrt(n))` is **absolutely convergent**
since `sum |a_n| ` as `sum_(n=1)^oo |(-1)^(n+1)/(nsqrt(n))|` is
convergent.

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.