Answer
The series is divergent and $b_n$ is not decreasing so, we cannot apply Alternating Series test.
Work Step by Step
Given: The series $\Sigma_{n=1}^\infty (-1)^{n+1} b_n$
$b_n=\dfrac{1}{n} \to $ Odd values and $b_n=\dfrac{-1}{n^2} \to$ Even values
However, $\dfrac{-1}{n^2}\lt \dfrac{-1}{n+1}$, for every $\dfrac{1}{n^2}$ term is less than the absolute value.
This means that $b_n$ is not decreasing so, Alternating Series test cannot be applied.
Also, $\Sigma_{n=1}^\infty (-1)^{n+1} b_n=\Sigma_{n=1}^\infty \dfrac{1}{2n-1}+\Sigma_{n=1}^\infty\dfrac{-1}{(2n)^2}$
and $\Sigma_{n=1}^\infty \dfrac{1}{2n-1} \gt \Sigma_{n=1}^\infty\dfrac{-1}{(2n)^2}$
The series $\Sigma_{n=1}^\infty \dfrac{1}{2n-1}$ diverges by the comparison test.
Thus, the series is divergent and $b_n$ is not decreasing so, we cannot apply Alternating Series test.
