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p-Series Convergence Theorem

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^p}} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + ~~ . . . }\)

\( p>1\)

\( 0 \lt p \leq 1\)

converges

diverges

The p-series is a pretty straight-forward series to understand and use. One detail you need to notice is that the theorem covers cases only where \( p > 0\).

Question: Do you know what happens when \( p \leq 0\)?

Okay, so testing for convergence and divergence of p-series looks pretty easy and, fortunately, it is. But it is also very powerful. To help you cement the idea more firmly in your mind, read through the panel below discussing the proof of this test. You do not have to understand the integral test at this point to understand this proof.

p-Series Convergence Theorem Proof

p-Series Convergence Theorem

The p-series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^p}}}\)

converges when

diverges when

\( p>1\)

\( 0 \lt p \leq 1\)

To prove this, we will use the Integral Test and the Special Improper Integral.

The integral test tells us that we can set up the integral \(\displaystyle{ \int_1^{\infty}{\frac{1}{x^p} dx} }\)

If this integral converges, then so does the series. Similarly, if the integral diverges, the series also diverges.

We will look at these five cases.
1. \( p > 1 \)
2. \( p = 1 \)
3. \( 0 \lt p \leq 1\)
4. \( p = 0 \)
5. \( p \lt 0 \)
Note: Although the last two cases are not part of the theorem, we will show what happens in those two cases to answer the question at the top of the page and for completeness.

Let's use the Integral Test to prove convergence and divergence by calculating the corresponding improper integrals. For most of the cases, we need to set up a limit as follows

\(\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x^p} dx}} }\)

Since the function \(\displaystyle{ \frac{1}{x^p} }\) is continuous on the interval \( [1,b] \), this integral can be evaluated. So, now let's look at each case individually.

Case 1: \( p > 1 \)
When \( p > 1 \), the integral is continuous and decreasing on the interval. So the Integral Test applies.

\(\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x^p} dx}} }\)

\(\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{ x^{-p} dx}} }\)

\(\displaystyle{ \lim_{b \to \infty}{ \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{b} } }\)

\(\displaystyle{ \frac{1}{-p+1} \lim_{b \to \infty}{\left[ b^{-p+1} - 1^{-p+1} \right]} }\)

\(\displaystyle{ \frac{1}{-p+1} [ 0 - 1 ] = \frac{1}{p-1} }\)

Since \(\displaystyle{ \frac{1}{p-1} }\) is finite, the integral converges and, therefore, by the Integral Test, the series also converges.

Case 2: \( p=1 \)
When \(p = 1\), we have

\(\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x} dx}} \to \lim_{b \to \infty }{ \left[ \ln(x) \right]_{1}^{b} } \to \lim_{b \to \infty }{ \ln(b) } - 0 = \infty }\)
Since the limit is infinity, the series diverges.

Case 3: \( 0 \lt p \lt 1\)
We will use the Integral Test again.

\(\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x^p} dx}} }\)

\(\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{ x^{-p} dx}} }\)

\(\displaystyle{ \lim_{b \to \infty}{ \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{b} } }\)

\(\displaystyle{ \frac{1}{-p+1} \lim_{b \to \infty}{\left[ b^{-p+1} - 1^{-p+1} \right]} }\)

So far in these calculations, we have the same equation as we did in case 1. However, in this case, \( 0 \lt p \lt 1\), which means that the exponent \(-p+1 > 0\) and therefore \(\displaystyle{ \lim_{b \to \infty}{ b^{-p+1} } \to \infty }\). So the entire integral diverges, which means that the series diverges.

Case 4: \( p = 0 \)
When \( p = 0 \), the series is \(\displaystyle{ \sum_{n=1}^{\infty}{1} }\). Since \(\displaystyle{ \lim_{n \to \infty}{1} = 1 \neq 0 }\) the divergence test tells us that the series diverges.

Case 5: \( p \lt 0 \)
We can write the fraction \(\displaystyle{ \frac{1}{n^p} = n^{-p} }\). Since \( p \lt 0 \), the exponent here is positive and so the terms are increasing. Since \(\displaystyle{ \lim_{n \to \infty}{ \frac{1}{n^p} } \neq 0 }\), the series diverges by the divergence test.

In Summary: For the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^p}}}\)
1. \( p > 1 \) converges by the integral test
2. \( p = 1 \) diverges by the integral test
3. \( 0 \lt p \lt 1\) diverges by the integral test
4. \( p = 0 \) diverges by the divergence test
5. \( p \lt 0 \) diverges by the divergence test

- qed -

Euler's Constant

Here is a great video discussing one use of a p-series.
Note - - Euler's Constant is not the same as Euler's Number.

PatrickJMT - Euler's Constant [16min-10secs]

video by PatrickJMT

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, determine whether these series converge or diverge using the p-series convergence theorem.

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\)

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\) diverges by the p-series test or the direct comparison test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\)

Solution

There are at least two ways to think about this one.
p-Series
First, if you factor out the constant in the numerator, you get \(\displaystyle{ 2 \sum_{n=1}^{\infty}{ \frac{1}{n} } }\) and the remaining sum is just a p-series with \(p=1\). Therefore, the series diverges.

Direct Comparison Test
If you didn't pick up the fact that you have p-series, you can use the Direct Comparison Test, comparing the given series with \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }\) which is a divergent p-series.
Now, to confirm that the given series diverges, we need to test \(\displaystyle{ \frac{2}{n} \geq \frac{1}{n} }\) for all n. If we multiply both sides of the inequality by n (which we can do since n is always positive), we get \( 2 \geq 1 \). This is true for all n, therefore the series diverges.

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\) diverges by the p-series test or the direct comparison test.

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\)

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) converges by the p-series test or the integral test.

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\)

Solution

This is a p-series with \( p=2 > 1\), so the series converges by the p-series test.
We could also have used the Integral Test, as follows.

\(\displaystyle{ \int_{1}^{\infty}{\frac{1}{x^2}dx} }\)

\(\displaystyle{ \lim_{b \to \infty}{\int_{1}^{b}{\frac{1}{x^2}dx}} }\)

\(\displaystyle{ \lim_{b \to \infty}{\int_{1}^{b}{x^{-2}dx}} }\)

\(\displaystyle{ \lim_{b \to \infty}{ \left[ -x^{-1} \right]_{1}^{b}} }\)

\(\displaystyle{ \lim_{b \to \infty}{-b^{-1} + 1^{-1}} }\)

\( 0 + 1 = 1 \)

Since the improper integral is finite, the series converges by the Integral Test.
Note: The value \(1\), from the integral is NOT necessarily what the series converges to. The significance of this number is only that it is finite.

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) converges by the p-series test or the integral test.

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\)

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\) is a convergent p-series.

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\)

Solution

253 solution video

video by PatrickJMT

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\) is a convergent p-series.

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\)

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\)

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\) is a divergent p-series.

Problem Statement

\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\)

Solution

254 solution video

video by PatrickJMT

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\) is a divergent p-series.

close solution

\(\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }\)

Problem Statement

\(\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }\)

Final Answer

The series \(\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }\) is a convergent p-series.

Problem Statement

\(\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }\)

Solution

255 solution video

video by PatrickJMT

Final Answer

The series \(\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }\) is a convergent p-series.

close solution
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