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You CAN Ace Calculus

17calculus > infinite series > p-series

 derivatives integrals basics of infinite series For the proof, you will also need these topics. improper integrals integral test

### Calculus Main Topics

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

Infinite Series - p-Series

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 $$p > 0$$. 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

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

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 -

### 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]

Practice Problems

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

 Level A - Basic

Practice A01

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n}{n^3} } }$$

solution

Practice A02

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

solution

Practice A03

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

solution

Practice A04

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

solution

Practice A05

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

solution

Practice A06

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

solution

10