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`sum_(n=2)^oo (-1)^n/(nlnn)` Determine whether the series converges absolutely or conditionally, or diverges.

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To determine the convergence or divergence of the series `sum_(n=2)^oo (-1)^n/(nln(n))` , we may apply Alternating Series Test.

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

1) `a_n` is monotone and decreasing sequence.

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

3) `a_ngt=0`

For the series `sum_(n=2)^oo (-1)^n/(nln(n))` , we have:

`a_n = 1/(nln(n))` which is a positive, continuous, and decreasing sequence from `N=2.`

Note: As "`n` " increases, the `nln(n)` increases then `1/(nln(n))` decreases.

Then, we set-up the limit as :

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

By alternating series test criteria, the series` sum_(n=2)^oo (-1)^n/(nln(n))`  converges.

The series `sum_(n=2)^oo (-1)^n/(nln(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=2)^oo |(-1)^n/(nln(n))|=sum_(n=2)^oo 1/(nln(n))`

Applying integral test for convergence, we evaluate the series as:

`int_2^oo1/(nln(n))dn=lim_(n-gtoo) int_2^t 1/(nln(n))dn`

Apply u-substitution: `u =ln(n)` then `du =1/ndn` .

`int 1/(nln(n))dn =int 1/(ln(n))*1/ndn `

                       `=int 1/u du`

                       ` =ln|u|`

Plug-in `u=ln(n)` on the indefinite integral `ln|u|` , we get:

`int_2^t 1/(nln(n))dn =ln|ln(n)||_2^t`

Applying definite integral formula: `F(x)|_a^b = F(b)-F(a)` .

`ln|ln(n)||_2^t =ln|ln(t)|-ln|ln(2)|`

Then, the limit becomes:

`lim_(n-gtoo) int_2^t1/(nln(n))dn =lim_(n-gtoo) [ln|ln(t)|-ln|ln(2)|]`


                                   `= oo - ln|ln(2` )|


`int_2^oo1/(nln(n))dn=oo` implies the series  `sum_(n=2)^oo |(-1)^n/(nln(n))|` diverges.



The series` sum_(n=2)^oo (-1)^n/(nln(n)) ` is conditionally convergent since`sum |a_n|` as   `sum_(n=2)^oo |(-1)^n/(nln(n))|` is divergent and `sum a_n` as` sum_(n=2)^oo (-1)^n/(nln(n))` is convergent.

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