$p$ -Series

Summary

The $p$ -series $\sum _{n=1}^{\mathrm{\infty }}1/n^p$ converges if and only if $p>1$ , and diverges if $p\le 1$ . The case $p=1$ is the harmonic series. The term test alone cannot prove divergence when $1/n^p\to 0$ (which holds for all $p>0$ ).

Prerequisites

Harmonic Series, Integral Test

Theorem

For real $p$ ,

Proof sketch: integral test with $f(x)=x^{-p}$ on $[1,\mathrm{\infty })$ .

Worked Example

• $\sum 1/n^2$ converges; in fact $\sum _{n=1}^{\mathrm{\infty }}1/n^2={\pi }^2/6=\zeta (2)$ .
• $\sum 1/\sqrt{n}$ diverges ( $p=1/2\le 1$ ), even though terms $\to 0$ .
• $\sum n=\sum n^{-(-1)}$ diverges ( $p=-1\le 1$ ).

Common Mistakes

• Claiming divergence of $\sum 1/n^p$ for $p>0$ solely because “terms do not go to zero”—they do go to zero for $p>0$ ; divergence for $0<p\le 1$ needs integral/comparison tests.
• Saying the series converges for $p\ge 1$ (false at $p=1$ ).

Connections

• Integral Test, Comparison Tests, Harmonic Series

References

$p$ -series are classified via the integral test in OpenStax Calculus Volume 2.^[1]