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`sum_(n=1)^oo (-1)^(n+1)/n^2` 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=1)^oo (-1)^(n+1)/n^2` , we may apply Alternating Series Test.

In Alternating Series Test, the series `sum (-1)^(n+1)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=1)^oo (-1)^(n+1)/n^2` , we have:

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

As "`n` " increases, the `1/n^2 ` decreases.

Then, we set-up the limit as :

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

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

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



The series `sum_(n=1)^oo (-1)^(n+1)/n^2` is absolutely convergent since `sum |a_n|` as `sum_(n=1)^oo |(-1)^(n+1)/n^2|` is convergent.

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