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Infinite Series - p-Series |
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p-Series Convergence Theorem |
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\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^p}} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + ~~ . . . }\) |
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\( p>1\) | | \( 0 \lt p \leq 1\) |
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converges | | diverges |
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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 \( p > 0\). Do you know what happens when \( p \leq 0\)? [ answer ]
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 Proof
p-Series Convergence Theorem |
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The p-series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^p}}}\)
converges when | | diverges when |
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\( p>1\) | | \( 0 \lt p \leq 1\) |
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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.
\(\displaystyle{
\begin{array}{rcl}
\lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x^p} dx}} & = & \lim_{b \to \infty}{ \int_1^{b}{ x^{-p} dx}} \\
& = & \lim_{b \to \infty}{ \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{b} } \\
& = & \frac{1}{-p+1} \lim_{b \to \infty}{\left[ b^{-p+1} - 1^{-p+1} \right]} \\
& = & \frac{1}{-p+1} [ 0 - 1 ] = \frac{1}{p-1}
\end{array}
}\)
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
\(\displaystyle{
\begin{array}{rcl}
\lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x} dx}} & = & \lim_{b \to \infty }{ \left[ \ln(x) \right]_{1}^{b} } \\
& = & \lim_{b \to \infty }{ \ln(b) } - 0 = \infty
\end{array}
}\)
Since the limit is infinity, the series diverges.
Case 3: \( 0 \lt p \lt 1\)
We will use the Integral Test again.
\(\displaystyle{
\begin{array}{rcl}
\lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x^p} dx}} & = & \lim_{b \to \infty}{ \int_1^{b}{ x^{-p} dx}} \\
& = & \lim_{b \to \infty}{ \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{b} } \\
& = & \frac{1}{-p+1} \lim_{b \to \infty}{\left[ b^{-p+1} - 1^{-p+1} \right]}
\end{array}
}\)
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 nth-term 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 nth-term 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 nth-term test
5. \( p \lt 0 \) diverges by the nth-term test
- qed -
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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] |
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Instructions - - Unless otherwise instructed, determine whether these series converge or diverge using the p-series test.
Practice A01 |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n}{n^3} } }\) |
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The series converges by the p-series test. |
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Starting with the nth-term test, we need to determine the limit
\(\displaystyle{ \lim_{n \to \infty}{ \frac{5n}{n^3} } }\).
There are two main ways to do this.
1. The first, and easiest way, is to use algebra.
2. The second way is to use L'Hôpital's Rule.
We will use algebra by multiplying the numerator and denominator by \(1/n\).
\(\displaystyle{
\begin{array}{rcl}
\lim_{n \to \infty}{ \frac{5n}{n^3} } & = & \lim_{n \to \infty}{ \frac{5n}{n^3} \frac{1/n}{1/n} } \\
& = & \lim_{n \to \infty}{ \frac{5}{n^2} } = 0
\end{array}
}\)
Since the limit is zero, we cannot say anything about whether this series converges or diverges using only the nth-term test.
By canceling the \(n\) in the numerator with one in the denominator ( which we do since the canceling operation does not change the domain ), we have \(a_n = 5/n^2\). This is a p-series with \(p=2>1\). So this series converges by the p-series test.
Note that even though the nth-term test cannot be used to determine convergence or divergence, we didn't just stop working the problem. The problem statement said to determine convergence or divergence. So stopping after applying the nth-term test would not be the completion of the problem. We actually needed to determine convergence or divergence. So once the first test failed, we moved on to another test until one of the tests determined convergence or divergence.
Practice A01 Final Answer |
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The series converges by the p-series test. |
Practice A02 |
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\(\displaystyle{\sum_{n=1}^{\infty}{\frac{2}{n}}}\) |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\) diverges by the p-series test or the direct comparison test. |
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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.
Practice A02 Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\) diverges by the p-series test or the direct comparison test. |
Practice A03 |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) |
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The p-series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) converges by the p-series test or the integral test. |
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This is a p-Series with \( p=2 > 1\), so the series converges.
We could also have used the Integral Test.
\(\displaystyle{
\begin{array}{rcl}
\int_{1}^{\infty}{\frac{1}{x^2}dx} & = & \lim_{b \to \infty}{\int_{1}^{b}{\frac{1}{x^2}dx}} \\\\
& = & \lim_{b \to \infty}{\int_{1}^{b}{x^{-2}dx}} \\\\
& = & \lim_{b \to \infty}{ \left[ -x^{-1} \right]_{1}^{b}} \\\\
& = & \lim_{b \to \infty}{-b^{-1} + 1^{-1}} \\\\
& = & 0 + 1 = 1
\end{array}
}\)
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.
Practice A03 Final Answer |
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The p-series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) converges by the p-series test or the integral test. |
Practice A04 |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\) |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\) is a convergent p-series. |
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Practice A04 Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\) is a convergent p-series. |
Practice A05 |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\) |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\) is a divergent p-series. |
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Practice A05 Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\) is a divergent p-series. |
Practice A06 |
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\(\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }\) |
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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. |
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Practice A06 Final Answer |
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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. |
