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.**

**Conclusion:**

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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