Using the p-series test to determine convergence

 
 
p-series test blog post.jpeg
 
 
 

What is the p-series test for convergence?

If we have a series ana_n in the form

an=n=11npa_n=\sum^{\infty}_{n=1}\frac{1}{n^p}

then we can use the p-series test for convergence to say whether or not ana_n will converge. The p-series test says that

ana_n will converge when p>1p>1

ana_n will diverge when p1p\le1

Krista King Math.jpg

Hi! I'm krista.

I create online courses to help you rock your math class. Read more.

 

The key is to make sure that the given series matches the format above for a p-series, and then to look at the value of pp to determine convergence.

 
 

How to use the p-series test to determine convergence?


 
Krista King Math Signup.png
 
Calculus 2 course.png

Take the course

Want to learn more about Calculus 2? I have a step-by-step course for that. :)

 
 

 
 

Let’s do a couple more examples where we determine convergence or divergence using the p-series test

Example

Use the p-series test to say whether or not the series converges.

n=11n\sum^{\infty}_{n=1}\frac{1}{\sqrt{n}}

In order to use the p-series test, we need to make sure the format of the given series matches the format above for a p-series, so we’ll rewrite the given series as

n=11n=n=11n12\sum^{\infty}_{n=1}\frac{1}{\sqrt{n}}=\sum^{\infty}_{n=1}\frac{1}{n^{\frac{1}{2}}}

In this format, we can see that p=1/2p=1/2. The p-series test tells us that ana_n diverges when p1p\le1, so we can say that this series diverges.


Let’s try a second example.


P-series test for Calculus 2.jpg

The key is to make sure that the given series matches the format above for a p-series, and then to look at the value of p to determine convergence.

Example

Use the p-series test to say whether or not the series converges.

n=11n43\sum_{n=1}^\infty\frac{1}{\sqrt[3]{n^4}}

In order to use the p-series test, we need to make sure the format of the given series matches the format above for a p-series, so we’ll rewrite the given series as

n=11n43=n=11(n4)13\sum_{n=1}^\infty\frac{1}{\sqrt[3]{n^4}}=\sum_{n=1}^\infty\frac{1}{(n^4)^\frac13}

n=11n43\sum_{n=1}^\infty\frac{1}{n^\frac43}

In this format, we can see that p=4/3p=4/3. The p-series test tells us that ana_n converges when p>1p>1, so we can say that this series converges.

 
Krista King.png
 

Get access to the complete Calculus 2 course