# The P-Series Test For Convergence

## P-Series

A **{eq}p {/eq}-series** is any one of a family of infinite series whose terms are reciprocal powers of natural numbers. The parameter {eq}p\in\mathbb{R} {/eq} specifies the power, which defines the series.

Any {eq}p {/eq}-series can be expressed in summation notation using the following **general formula**:

$$\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^p} $$

For example, if we choose the power {eq}p=2 {/eq}, the corresponding {eq}p {/eq}-series is the sum of all reciprocals of perfect square numbers:

$$\begin{eqnarray} \sum_{n=1}^\infty \dfrac{1}{n^2} &=& \dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \ldots \\ &=& 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \ldots \end{eqnarray} $$

{eq}p {/eq}-series in **calculus** has applications because series are closely connected to definite integrals, which compute the area underneath a function {eq}f(x) {/eq}. In particular, the convergence of some integrals can be determined by comparing the convergence of a corresponding series, which could be a {eq}p {/eq}-series.

### Harmonic Series

The series which consists of the sum of the reciprocals of all natural numbers is known as the **harmonic series**. Its general formula, and calculation term-by-term, are

$$\displaystyle \sum_{n=1}^\infty \dfrac{1}{n } = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \ldots $$

We can notice that this is simply a special case of a {eq}p {/eq}-series with the power {eq}p=1 {/eq}. This simple member of the family turns out to play a key role in determining the {eq}p {/eq}-series convergence of divergence, as we'll see momentarily, when we discuss the **{eq}p {/eq}-series test**.

### Sum of P-Series

There is no easy way to calculate the exact sum of a given {eq}p {/eq}-series, though the value of some specific series have been determined. For example the exact value of the series for {eq}p=2 {/eq} is

$$\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{ \pi^2}{6} \approx 1.65 $$

The {eq}p {/eq}-series test can at least determine whether or not a particular series converges at all, and an approximate value can then be found by adding up terms in the sum.

## P-Series Test for Convergence

Whether or not a {eq}p {/eq}-series converges to a specific value can be determined using the very simple **{eq}p {/eq}-series test for convergence**. The test states that the series

$$\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^p} $$

will **converge if {eq}p>1 {/eq}**, otherwise it will diverge.

This rule for {eq}p {/eq} series is consistent with our earlier example with power {eq}p =2 > 1 {/eq}, which we saw was a convergent series.

## P-Series Examples

Lets consider several {eq}p {/eq}-series examples, and determine their convergence using the {eq}p {/eq}-series test.

### Example 1

Lets begin by determining the convergence of the harmonic series. Since this is a {eq}p {/eq}-series with power {eq}p=1 {/eq}, the {eq}p {/eq}-series test tells us that the harmonic series **diverges**.

The terms of the harmonic series quickly become very small numbers. If you try adding them up, it may seem that the sum is going to converge. Nevertheless, as more terms are added the sum will grow very large - but very slowly!

According to the {eq}p {/eq}-series rule, the harmonic series is the dividing line between convergent and divergent {eq}p {/eq}-series, since the series *do *converge for any power strictly above {eq}p=1 {/eq}.

### Example 2

Does {eq}\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt n} {/eq} converge?

If we recall that {eq}\sqrt n = n^{\frac{1}{2} } {/eq} then we can recognize this series a a {eq}p {/eq}-series with power {eq}p=\frac{1}{2} {/eq}. Since {eq}p<1 {/eq} this series must **diverge**.

### Example 3

Does {eq}\displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt[3] {n^4 \ } } {/eq} converge?

Similarly, according to the rules of roots and exponents, we can recognize that

$$\dfrac{1}{\sqrt[3] {n^4 \ } } = \dfrac{1}{n^{\frac{4}{3} } } $$

making this a {eq}p {/eq}-series with power {eq}p=\frac{4}{3} > 1 {/eq}. According to the test, this series **converges**.

### Example 4

Does {eq}\displaystyle \sum_{n=1}^\infty \dfrac{ 1+ n }{ n^2 } {/eq} converge?

We can separate this sum into two distinct series, and analyze the convergence of each one individually. We have

$$\displaystyle \sum_{n=1}^\infty \dfrac{ 1 +n }{ n^2 } \ = \ \sum_{n=1}^\infty \dfrac{ 1 }{ n^2 } + \sum_{n=1}^\infty \dfrac{ n }{ n^2 } \ = \ \sum_{n=1}^\infty \dfrac{ 1 }{ n^2 } + \sum_{n=1}^\infty \dfrac{ 1 }{ n } $$

Now we can see that this series is actually the sum of two different {eq}p {/eq}-series. The first converges since its power is {eq}p=2>1 {/eq}, but the second is the divergent harmonic series, with {eq}p=1 {/eq}.

Since one part of the series is divergent, the overall series also **diverges**.

## Lesson Summary

In this lesson we've learned how to recognize **{eq}p {/eq}-series**, whose terms are reciprocal powers of natural numbers. Determining the convergence of series of this type has applications in the field of **calculus**.

All {eq}p {/eq}-series can expressed in the **general form**

$$\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^p} $$

for some fixed power {eq}p\in \mathbb{R} {/eq}. The special case of {eq}p=1 {/eq} is also known as the **harmonic series**. The convergence or divergence of a {eq}p {/eq}-series can be determined using the **{eq}p {/eq}-series test**. If the power in the denominator satisfies {eq}p>1 {/eq} then the series converges, otherwise it diverges. The convergence test can be applied to a a variety of series is we recall to express roots as powers, and to decompose rational expressions into separate terms where possible.

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#### What is the P series formula?

All *p*-series can be represented as the infinite sum of the reciprocal powers of the natural numbers. The general formula is the sum of *1/n^p* for *n* between 1 and infinity.

#### What is a convergent p series?

An infinite series converges if the limit of its sum approaches a specific finite value. *p*-series converge when the power appearing in the denominator of each term satisfies *p>1*.

#### What is the P series test?

The p-series test can be used to determine if a *p*-series converges or diverges. It converges if, and only if, the power satisfies *p>1*.

#### How do you know if a series is P series?

The terms in a *p*-series are reciprocal powers of natural numbers. This means they have the form *1/n^p* for some fixed power *p*.

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