Determining Whether a Series Is Absolutely or Conditionally Convergent or Divergent

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 \(\lim\limits_{n\,\rightarrow \infty }\,a_{n}\):

\(\lim\limits_{n\,\rightarrow \infty }\dfrac{1}{n^{p}}=0\quad 0<p≤1\)

\(\lim\limits_{n\,\rightarrow \infty }\dfrac{1}{n^{p}}=1\quad p=0\)

\(\lim\limits_{n\,\rightarrow \infty }\dfrac{1}{n^{p}}= \infty \quad p<0\)

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 \(p\leq 0.\)