20. Convergence of Positive Series

b.2. The \(p\)-Series

The series \(\displaystyle \sum_{n=n_o}^\infty \dfrac{1}{n^p}\) is called a \(p\)-series.
The harmonic series \(\displaystyle \sum_{n=n_o}^\infty \dfrac{1}{n}\) is the special case of the \(p\)-series with \(p=1\).

The \(p\)-series \(\displaystyle \sum_{n=n_o}^\infty \dfrac{1}{n^p}\) is convergent if \(p \gt 1\) and is divergent if \(p \le 1\).
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We check the different values of \(p\).

  1. The harmonic series with \(p=1\) was shown to be divergent in an exercise at the bottom of the previous page.
  2. For \(p \le 0\), the series \(\displaystyle \sum_{n=n_o}^\infty \dfrac{1}{n^p} = \sum_{n=n_o}^\infty n^{-p}\) is divergent by the \(n^\text{th}\)-Term Divergence Test.
  3. For \(p \gt 0\) but \(p \ne 1\), we apply the Integral Test:
      The function \(f(x)=\dfrac{1}{x^p}\) is continuous, positive and decreasing. Further, \[\begin{aligned} \int_{n_o}^\infty \dfrac{1}{x^p}\,dx &=\int_{n_o}^\infty x^{-p}\,dx =\left[\dfrac{x^{-p+1}}{-p+1}\right]_{n_o}^\infty \\ &=\lim_{b\to\infty}\left[\dfrac{b^{-p+1}}{-p+1}\right] -\left[\dfrac{n_o^{-p+1}}{-p+1}\right] \end{aligned}\]
    1. If \(0< p< 1\), then \(-p+1>0\) and \(\displaystyle \lim_{b\to\infty}\left[\dfrac{b^{-p+1}}{-p+1}\right] =\infty\).
      Consequently, \(\displaystyle \int_{n_o}^\infty \dfrac{1}{x^p}\,dx\) and hence \(\displaystyle \sum_{n=n_o}^\infty \dfrac{1}{n^p}\) are divergent.
    2. If \(p>1\), then \(-p+1< 0\) and \(\displaystyle \lim_{b\to\infty}\left[\dfrac{b^{-p+1}}{-p+1}\right] =0\).
      Consequently, \(\displaystyle \int_{n_o}^\infty \dfrac{1}{x^p}\,dx\) and hence \(\displaystyle \sum_{n=n_o}^\infty \dfrac{1}{n^p}\) are convergent.

Determine if \(\displaystyle \sum_{n=1}^\infty \dfrac{3}{n^2}\) is convergent or divergent.

You can apply the Integral Test directly, or simply say \(\displaystyle \sum_{n=1}^\infty \dfrac{3}{n^2}\) is \(3\) times a \(p\)-series with \(p=2 \gt 1\). Consequently, \(\displaystyle \sum_{n=1}^\infty \dfrac{3}{n^2}\) is convergent.

Determine if \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptsize3]{n}}\) is convergent or divergent.

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptsize3]{n}}\) is divergent.

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptsize3]{n}}\) is a \(p\)-series with \(p=\dfrac{1}{3}<1\). So it is divergent.

Determine if \(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptsize3]{n^4}}\) is convergent or divergent.

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptsize3]{n^4}}\) is convergent.

\(\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[\scriptsize3]{n^4}}\) is a \(p\)-series with \(p=\dfrac{4}{3} \gt 1\). So it is convergent.

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