Determine the numbers $p$ for which the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{(-1)^{k+1}}{k^{p\,}}$ is absolutely convergent, conditionally convergent, or divergent.

Solution We begin by testing the series for absolute convergence. The series of absolute values is $\sum\limits_{k\,=\,1}^{\infty}\left\vert \dfrac{(-1)^{k+1}}{k^{p\,}}\right\vert=\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{p\,}}$. This is a $p$-series, which converges if $p>1$ and diverges if $p\leq 1$. So, $\sum\limits_{k\,=\,1}^{\infty }\dfrac{(-1)^{k+1}}{k^{p\,}}$ is absolutely convergent if $p>1$.

It remains to determine what happens when $p\leq 1$. We use the Alternating Series Test, and begin by investigating $\lim\limits_{n\,\rightarrow \infty }\,a_{n}$:

Consequently, $\sum\limits_{k\,=\,1}^{\infty }\dfrac{(-1)^{k+1}}{k^{p\,}}$ diverges if $p\leq 0$.

Continuing, we check the second condition of the Alternating Series Test when $0<p\leq 1.$ Using the related function $f(x) =\dfrac{1}{x^{p}},$ for $x>0,$ we have $f^{\prime} (x) =-\dfrac{p}{x^{p+1}}$. Since $f^{\prime} (x) <0$ for $0<p\leq 1$, the second condition of the Alternating Series Test is satisfied. We conclude that $\sum\limits_{k\,=\,1}^{\infty }\dfrac{(-1)^{k+1}}{k^{p\,}}$ is conditionally convergent if $0<p\leq 1$.

To summarize, the alternating $p$-series, $\sum\limits_{k\,=\,1}^{\infty }\dfrac{(-1)^{k+1}}{k^{p\,}}$ is absolutely convergent if $p>1,$ conditionally convergent if $0<p\leq 1,$ and divergent if $p\leq 0.$