Answer
The series is divergent and $b_n$ is not decreasing so we cannot apply the Alternating Series test.
Work Step by Step
Given: The series $\Sigma_{n=1}^\infty (-1)^{n+1} b_n$
Here, we have $b_n=\dfrac{1}{n}$ for odd values and $b_n=\dfrac{-1}{n^2}$ for even values.
We can see that $\dfrac{-1}{n^2}\lt \dfrac{-1}{n+1}$, for every $\dfrac{1}{n^2}$ term is less than the absolute value.
Thus, we can conclude that $b_n$ is not decreasing so we cannot apply the Alternating Series test.
$\Sigma_{n=1}^\infty (-1)^{n+1} b_n=\Sigma_{n=1}^\infty \dfrac{1}{2n-1}+\Sigma_{n=1}^\infty\dfrac{-1}{(2n)^2}$
Here, $\Sigma_{n=1}^\infty \dfrac{1}{2n-1} \gt \Sigma_{n=1}^\infty\dfrac{-1}{(2n)^2}$
Thus, $\Sigma_{n=1}^\infty \dfrac{1}{2n-1}$ diverges by the comparison test.
Hence, it has been proved that the series is divergent and $b_n$ is not decreasing so we cannot apply the Alternating Series test.
