(S5) p-Series#

In this lesson we are going to see:

  • the definition of a \(p\)-series

  • how to determine if a \(p\)-series is convergent or divergent.

Review Videos#

Definition#

A \(p\) series is just a special type of series where we essentially try to add up the terms in a sequence formed from a power function.

\(p\) Series

A \(p\) series is a series of the form:

\[ \sum_{n=1}^{\infty} \dfrac{1}{n^p}= \dfrac{1}{1^p}+\dfrac{1}{2^p}+\dfrac{1}{3^p} + \cdots + \dfrac{1}{n^p}+ \cdots \]

Special Case: We sometimes call the case where \(p=1\) a harmonic series.

Comparison to Geometric Series#

\(p\) Series
\[ \sum_{n=1}^{\infty} \dfrac{1}{n^p} \]

  • \(n\) is in the base

  • the exponent is a constant

Geometric Series
\[ \sum_{n=0}^{\infty} ar^n \]

  • \(n\) is in the exponent

  • the base is a constant

Convergence#

Convergence of a \(p\) Series

The convergence of a \(p\)-series depends on the value of \(p\):

\[ \sum_{n=1}^{\infty} \dfrac{1}{n^p} \]

Specifically, this series is:

  • convergent when \(p>1\)

  • divergent when \(p\leq 1\)

Tip

We can relax the starting index value to include other values besides \(n=1\), and this convergence result still applies.

Sum#

Unlike geometric series, there is no nice general formula for the sum of a \(p\)-series.

There are some special cases where we actually do know the sum, like below. But in general for \(p\) series, we normally only state whether it is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{1}{n^2}=\dfrac{\pi^2}{6} \]

Improper Integrals?#

Compare the convergence of a \(p\) series, with what we saw for \(p\)-form improper integrals.

Improper Integrals (\(p\)-forms)

The convergence of the \(p\)-form improper integral depends on the value of \(p\):

\[ \int_1^{\infty} \dfrac{1}{x^p}\, dx \]

Specifically, this improper integral is:

  • convergent when \(p>1\)

  • divergent when \(p\leq 1\)

Is this coincidence?

No. As we’re going to see when we talk about the Integral Test, there is a general result that relates the convergence and divergence of improper integrals with the convergence and divergence of series (with some restrictions).

Example 1#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{1}{n^2} \]
Solution (Click to see the steps.)

This is a \(p\) series with \(p=2\), which is greater than 1.

The series is therefore convergent.

Example 2#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}} \]
Solution (Click to see the steps.)

This is a \(p\) series with \(p=\tfrac{1}{2}\) which is less than 1.

The series is therefore divergent.

Example 3#

Determine if the following series is convergent or divergent.

\[ \sum_{n=5}^{\infty} \dfrac{1}{n} \]
Solution (Click to see the steps.)

This is a \(p\) series with \(p=1\) and is therefore divergent.