`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.
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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