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P Series

Q: What is p-series test for convergence?

If you are asking when the p-series converges and when it diverges, the p-series converges for p>1 and diverges otherwise.

Q: How do you find p-series?

A p-series is of the form ∑ (1/n) p , but sometimes you may have to do some algebra on the general term to verify if it is of the form (1/n) p .

Suppose each time you buy fast food you get a ticket, and if you collect all of the different kinds of tickets you will more fast food. If there are different types of tickets, how many times do you need to buy fast food before you can expect to win a prize? This is called the Coupon Collector's Problem, and it is an application of the harmonic series.

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Fact Checked Content
Last Updated: 23.09.2022
5 min reading time

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This article will explore a generalization of the harmonic series, p-series, its definition or sum, criterion of convergence, and other related convergence tests.

p-Series Sum

So first of all, what is a harmonic series? It is a kind of p-series. Now you need to know what a p-series is.

A p-series is a series of the form

$\sum_{n = 1}^{\infty} {(\frac{1}{n})}^{p} = \sum_{n = 1}^{\infty} \frac{1}{n^{p}}$

where $p$ is a real number.

When this is called the Harmonic series.

Decide if the series

$\sum_{n = 1}^{\infty} \frac{\sqrt[4]{n}}{n^{5}}$

is a p-series or not.

Answer:

It doesn't look like a p-series at first glance, but let's do some algebra just to be sure. Here

$\frac{\sqrt[4]{n}}{n^{5}} = \frac{n^{\frac{1}{4}}}{n^{5}} = \frac{1}{n^{5} \cdot n^{- \frac{1}{4}}} = \frac{1}{n^{5 - \frac{1}{4}}} = \frac{1}{n^{\frac{19}{4}}} = {(\frac{1}{n})}^{\frac{19}{4}}$

so in fact this is a p-series.

Is the series

$\sum_{n = 1}^{\infty} \frac{1}{3^{n}}$

a p-series?

Answer:

No, because in a p-series it needs to be $\frac{1}{n}$ raised to a constant power, not a constant raised to the $n$ power. In fact this series also has a special name, it is called a Geometric series. For more information on this type of series see Geometric Series.

p-Series and the Integral Test

For more information on why the Integral Test works and how to use it, see Integral Test. Let's look at an example of applying the Integral Test to the p-series.

Does the series

$\sum_{n = 1}^{\infty} \frac{1}{n \cdot \sqrt[3]{n}}$

converge or diverge?

Answer:

While this may not look like a p-series, remember that

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

so this is actually a p-series. To use the Integral Test, take the function

$f (x) = \frac{1}{x^{\frac{4}{3}}}$ .

This function is decreasing, continuous, and positive for which means the conditions are met to apply the Integral Test. Then integrating,

$\int_{1}^{\infty} f (x) d x = \int_{1}^{\infty} \frac{1}{x^{\frac{4}{3}}} d x = \int_{1}^{\infty} x^{- \frac{4}{3}} d x = \lim_{k \to \infty} \int_{1}^{k} x^{- \frac{4}{3}} d x = \lim_{k \to \infty} [{- 3 x^{- \frac{1}{3}}}_{1}^{k}] = \lim_{k \to \infty} [- 3 k^{- \frac{1}{3}} - (- 3)] = 3 .$

Since the integral converges, by the Integral Test the series converges too.

p-Series Convergence

Figuring out when a general p-series converges and diverges is an application of the Integral Test as well.

Proof of p-Series Convergence

To prove whether or not the p-series converges or diverges, you would use the Integral Test exactly as in the example above, but with a general value for $p$ rather than the $p = \frac{4}{3}$ used in the example. You can state the results of that Integral Test as:

Using the Integral Test you can see that:

If $p > 1$ , the p-series converges,
If $p \leq 1$ , the p-series diverges.

Sometimes the information in the Deep Dive above is called the p-Series Test, even though it is really just properties of the p-Series and not a real test.

This means that the Harmonic series diverges. How you can use this in the Coupon Collector's Problem is beyond what this article will cover, but if there are 50 different kinds of tickets you can expect to purchase fast food about 225 different times to collect all of the tickets, and that is assuming that there aren't any rare ones!

Does the series

$\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$

converge or diverge?

Answer:

In this example $p = 2$ . Since $p > 1$ , this series converges.

Explain why the series $\sum_{n = 1}^{\infty} {(\frac{1}{n})}^{- 2}$ diverges by writing out some of the partial sums and showing what happens.

Answer:

Writing out the partial sums,

$\begin{matrix} s_{1} & = & {(\frac{1}{1})}^{- 2} = 1 \\ s_{2} & = & s_{1} + {(\frac{1}{2})}^{- 2} = 1 + 4 = 5 \\ s_{3} & = & s_{2 +} {(\frac{1}{3})}^{- 2} = 5 + 9 = 14, \end{matrix}$

and as you can see they just keep getting bigger. That means the series diverges.

p-Series and the Comparison Test

In general the p-series isn't one you will actually find the sum of. Instead because you know exactly what makes it converge or diverge, it is useful in comparing other series to. For more information on tests for convergence see Convergence Tests, but remember that to use the Comparison Test you need:

a series with positive terms you can compare to, and
to be able to tell whether or or not your series with positive terms diverges or converges.

The p-series has positive terms, and it is easy to tell if it converges or diverges. So it is very helpful when trying to apply the Comparison Test.

Decide if the series

$\sum_{n = 1}^{\infty} \frac{2^{n}}{n^{2} 3^{n}}$

converges or diverges.

Answer:

First let's rewrite the series in question as

$\sum_{n = 1}^{\infty} {(\frac{2}{3})}^{n} \frac{1}{n^{2}}$ .

You already know that the series

$\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$

is a p-series with $p = 2$ , and it converges. It also has all positive terms. So if you can squash the terms of the series you are looking at (which also has all positive terms) under the terms of the p-series, you can use the Comparison Test to say that the new series converges.

Looking at

${(\frac{2}{3})}^{n}$ ,

because $n \geq 1$ you can say that

${(\frac{2}{3})}^{n} < 1$ .

${(\frac{2}{3})}^{n} {(\frac{1}{n})}^{2} < 1 \cdot {(\frac{1}{n})}^{2} = {(\frac{1}{n})}^{2}$ .

That means that by the Comparison Test using the p-series with , the series

$\sum_{n = 1}^{\infty} \frac{2^{n}}{n^{2} 3^{n}}$

converges.

p-Series - Key takeaways

A p-series is a series of the form
$\sum_{n = 1}^{\infty} \frac{1}{n^{p}}$
where is a real number.
When the p-series is called the Harmonic series.
If , the p-series converges.
If , the p-series diverges.

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Frequently Asked Questions about P Series

What is p-series test for convergence?

If you are asking when the p-series converges and when it diverges, the p-series converges for p>1 and diverges otherwise.

How do you find p-series?

A p-series is of the form ∑ (1/n)^p, but sometimes you may have to do some algebra on the general term to verify if it is of the form (1/n)^p.

When can the p-series test be used?

It depends on what you mean by the p-series test. If you just want to know when the p-series converges and when it diverges, the p-series converges for p>1 and diverges otherwise.

How do you prove p-series?

Well, you don't prove p-series. But if you want to prove when a p-series converges and when it diverges, you use the Integral Test.

What is the p-series?

A p-series is an infinite series where each term has the form (1/n)^p.

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