The p-Series ∑ 1/n^p is convergent if p is greater than 1

In this video, I use the Integral Test to show that the p-series, whose terms are 1/n^p, is convergent if p is greater than 1 and divergent for all other values of p. When p is less than 0, the terms approach infinity; when p = 0, the terms equal 1, hence it is divergent by the test for divergence; and when p = 1, we get the harmonic series, which is also divergent. I also illustrate this with examples on a convergent and a divergent p-series. Note also that when using the Integral Test, the corresponding improper integrals do NOT, in general, equal the sum of the series.

#math #calculus #series #pseries #education

Timestamps

• Example 2: The convergence of the p-Series – 0:00
  • Solution: If p is less than 0, the series diverges because the terms approach infinity – 0:20
  • If p = 0, the terms approach 1, thus the series diverges as well – 1:27
  • If p is greater than 1, the series is convergent by the Integral Test and diverges if p is less than or equal to 1 – 3:50
  • If p = 1, the series is the Harmonic series, and is divergent – 5:31
• To use the Integral Test, we need to find the antiderivative, which is often difficult or impossible, so we need other tests for convergence – 6:10
• Theorem 1: p-Series – 6:44
• Example 3: Convergent and divergent p-series – 7:14
• Note that the sum of the improper integral is generally NOT equal to the sum of the series – 8:30

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