You CAN Ace Calculus

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

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

### Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

17calculus > infinite series > p-series

 p-Series p-Series Convergence Theorem Proof Euler's Constant Practice

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 $$p \leq 0$$? Think about it for a minute and then click here for the 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.

 $$\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x^p} dx}} }$$ $$\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{ x^{-p} dx}} }$$ $$\displaystyle{ \lim_{b \to \infty}{ \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{b} } }$$ $$\displaystyle{ \frac{1}{-p+1} \lim_{b \to \infty}{\left[ b^{-p+1} - 1^{-p+1} \right]} }$$ $$\displaystyle{ \frac{1}{-p+1} [ 0 - 1 ] = \frac{1}{p-1} }$$

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

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

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.

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

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

Problem Statement

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

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.

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

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

Problem Statement

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

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

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

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

Problem Statement

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

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

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

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

Problem Statement

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

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

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.