pSeries Convergence Theorem  

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^p}} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + ~~ . . . }\)  
\( 0 \lt p \leq 1\) 
\( p>1\)  
diverges 
converges 
Note  In the proof section below, we show a more efficient way to write this theorem statement.
The pseries is a pretty straightforward 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 pseries 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.
pSeries Convergence Theorem 

The pseries \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^p}}}\)
diverges 
converges  

\( 0 \lt p \leq 1\) 
\( p>1\) 
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.
\(\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{\frac{1}{x^p} dx}} }\) 
\(\displaystyle{ \lim_{b \to \infty}{ \int_1^{b}{ x^{p} dx}} }\) 
\(\displaystyle{ \lim_{b \to \infty}{ \left[ \frac{x^{p+1}}{p+1} \right]_{1}^{b} } }\) 
\(\displaystyle{ \frac{1}{p+1} \lim_{b \to \infty}{\left[ b^{p+1}  1^{p+1} \right]} }\) 
\(\displaystyle{ \frac{1}{p+1} [ 0  1 ] = \frac{1}{p1} }\) 
Since \(\displaystyle{ \frac{1}{p1} }\) 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 
Note: We have also seen the pseries theorem where, instead of writing \( 0 \lt p \leq 1\) for where the series diverges, it says the series diverges for \( p \leq 1 \). Of course, this includes all of the cases above. This is a more efficient way to write the theorem statement.
Euler's Constant
Here is a great video discussing one use of a pseries.
Note   Euler's Constant is not the same as Euler's Number.
video by PatrickJMT 

Practice
Unless otherwise instructed, determine whether these series converge or diverge using the pseries convergence theorem.
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\)
Problem Statement 

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\) using the Direct Comparison Test, if possible. If the DCT is inconclusive, use another test to determine convergence or divergence. Make sure to specify what test you used in your final answer.
Final Answer 

The series diverges by the pseries test or the Direct Comparison Test.
Problem Statement 

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\) using the Direct Comparison Test, if possible. If the DCT is inconclusive, use another test to determine convergence or divergence. Make sure to specify what test you used in your final answer.
Solution 

There are at least two ways to think about this one.
pSeries
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 pseries with \(p=1\). Therefore, the series diverges.
Direct Comparison Test
If you didn't pick up the fact that you have pseries, you can use the Direct Comparison Test, comparing the given series with \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }\) which is a divergent pseries.
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 diverges by the pseries test or the Direct Comparison Test. 
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\)
Problem Statement 

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) using the integral test, if possible.
Final Answer 

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

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }\) using the integral test, if possible.
Solution 

This is a pseries with \( p=2 > 1\), so the series converges by the pseries 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 pseries test or the integral test. 
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^2} } }\)
Problem Statement 

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^2} } }\) converges or diverges using the pseries convergence theorem.
Solution 

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

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\) converges or diverges using the pseries convergence theorem.
Final Answer 

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

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\) converges or diverges using the pseries convergence theorem.
Solution 

video by PatrickJMT 

Final Answer 

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{1.4}} } }\) is a convergent pseries. 
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\)
Problem Statement 

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\) converges or diverges using the pseries convergence theorem.
Final Answer 

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

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\) converges or diverges using the pseries convergence theorem.
Solution 

video by PatrickJMT 

Final Answer 

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{0.7}} } }\) is a divergent pseries. 
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\(\displaystyle{ \sum_{n=1}^{\infty}{ n^{\pi} } }\)
Problem Statement 

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ n^{\pi} } }\) converges or diverges using the pseries convergence theorem.
Solution 

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\(\displaystyle{ \sum_{n=1}^{\infty}{ (n^{2.4} + 8n^{1.6}) } }\)
Problem Statement 

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ (n^{2.4} + 8n^{1.6}) } }\) converges or diverges using the pseries convergence theorem.
Solution 

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\(\displaystyle{ 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + . . . }\)
Problem Statement 

Determine whether the series \(\displaystyle{ 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + . . . }\) converges or diverges using the pseries convergence theorem.
Solution 

This problem is solved by two different instructors.
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\(\displaystyle{ 1 + \frac{1}{\sqrt[3]{4}} + \frac{1}{\sqrt[3]{9}} + \frac{1}{\sqrt[3]{16}} + . . . }\)
Problem Statement 

Determine whether the series \(\displaystyle{ 1 + \frac{1}{\sqrt[3]{4}} + \frac{1}{\sqrt[3]{9}} + \frac{1}{\sqrt[3]{16}} + . . . }\) converges or diverges using the pseries convergence theorem.
Solution 

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

Determine whether the series \(\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }\) converges or diverges using the pseries convergence theorem.
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 pseries.
Problem Statement 

Determine whether the series \(\displaystyle{ 1+ \frac{1}{2\sqrt[3]{2}} + \frac{1}{3\sqrt[3]{3}} + \frac{1}{4\sqrt[3]{4}} + . . . }\) converges or diverges using the pseries convergence theorem.
Solution 

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 pseries. 
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt[3]{n^4}} } }\)
Problem Statement 

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt[3]{n^4}} } }\) converges or diverges using the pseries convergence theorem.
Solution 

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \sqrt[5]{n^2} } }\)
Problem Statement 

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \sqrt[5]{n^2} } }\) converges or diverges using the pseries convergence theorem.
Solution 

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3}{\sqrt[3]{n^2}} } }\)
Problem Statement 

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3}{\sqrt[3]{n^2}} } }\) converges or diverges using the pseries convergence theorem.
Solution 

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3 \sqrt{n}} } }\)
Problem Statement 

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3 \sqrt{n}} } }\) converges or diverges using the pseries convergence theorem.
Solution 

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{4\sqrt{n}}{n^4} } }\)
Problem Statement 

Determine whether the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{4\sqrt{n}}{n^4} } }\) converges or diverges using the pseries convergence theorem.
Solution 

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You CAN Ace Calculus
For the proof, you will also need these topics. 
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Practice Instructions
Unless otherwise instructed, determine whether these series converge or diverge using the pseries convergence theorem.