Absolute and conditional convergence applies to all series whether a series has all positive terms or some positive and some negative terms (but the series is not required to be alternating).
One unique thing about series with positive and negative terms (including alternating series) is the question of absolute or conditional convergence. Once convergence of the series is established, then determining the convergence of the absolute value of the series tells you whether it converges absolutely or conditionally. Formally, here's what it looks like.
Definitions of Absolute and Conditional Convergence  

Given that \(\sum{a_n}\) is a convergent series.  
if \(\sum{\lefta_n\right}\) converges 
if \(\sum{\lefta_n\right}\) diverges  
then \(\sum{a_n}\) converges absolutely 
then \(\sum{a_n}\) converges conditionally 
Here is a table that summarizes these ideas a little differently.
\(\sum{a_n}\) 
\(\sum{\lefta_n\right}\) 
conclusion  

converges 
converges 
\(\sum{a_n}\) converges absolutely  
converges 
diverges 
\(\sum{a_n}\) converges conditionally 
Absolute Convergence Theorem
Absolute Convergence Theorem 

If the series \( \sum{\lefta_n\right} \) converges, 
Alternatively, it is possible to determine the convergence of the absolute value of the series first. Then, if the absolute value of the series converges, you can use the Absolute Convergence Theorem to say that the alternating series also converges and converges absolutely.
Additionally, if you have a series with some negative terms (but not all) and it is not an alternating series, you can use this theorem to determine convergence. Specifically, if the absolute value of the series converges, then the series will converge. Notice that this theorem says nothing about divergence, so you cannot make any assumption about convergence or divergence if this theorem does not hold.
Understanding The Theorem
Let's get an idea of why this theorem works. First, let's think about the series \( \sum{a_n} \) with positive and negative terms. As we add up the terms, we will be adding some positive terms and subtracting other terms (the negative ones). Compare that to the series \( \sum{\lefta_n\right} \). With this series, we add all the terms as we go, i.e. we add all the positive terms and then we take the absolute value of the negative terms before we add those terms also. So each partial sum of this series is greater than the partial sum of the previous series. Consequently, as we continue to calculate the partial sums, we can say that the second sum is larger than the first sum. So if the larger sum converges, the smaller sum also has to converge.
Here is a simple example that should give you an intuitive feel for this.
For this finite alternating series \( 1  2 + 3  4 + 5 = 3 \).
Now calculate the absolute value of that series to get \( 1 + 2 + 3 + 4 + 5 = 15 \)
If we are always adding the numbers and never subtracting, the sum will always be larger for the absolute value series. And in the case of an infinite sum, if the larger series converges, logically, the smaller one will too.
Note  This is not a formal proof of the theorem. It is just an example to give you a feel for it.
Okay, let's watch some videos to get a better idea of what is going on. This next video clip has a great discussion on absolute convergence including using some examples. Notice that he does not use the term conditional convergence. Instead, he just says that the series does not converge absolutely.
video by Dr Chris Tisdell 

Here is another short video clip explanation of absolute and conditional convergence.
video by PatrickJMT 

Freaky Consequence of Conditionally Convergent Infinite Series
As you probably know by now, when you start working with numbers out to infinity, strange things happen. This is certainly true of infinite series which are conditionally convergent. The strange thing is that, when you rearrange the sum, you can get different values to which the series converges, i.e. the commutative property of numbers does not hold! Whoa!
In fact, not only can you get different values, it is possible to rearrange a conditionally convergent infinite series in order to get any number we want, including zero and infinity! This is called the Riemann Series Theorem.
Want to know more? Check out this wikipedia page.
Okay, time for some practice problems. These problems are not alternating series. For practice problems applying absolute and conditional convergence to alternating series, see the alternating series page.
Practice
Unless otherwise instructed, for each of the following series:
1. Determine whether it converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{\sin(n)}{n^2} } }\)
Problem Statement 

For the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{\sin(n)}{n^2} } }\)
1. Determine whether the series converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.
Final Answer 

The series converges absolutely by the direct comparison test and the absolute convergence theorem.
Problem Statement 

For the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{\sin(n)}{n^2} } }\)
1. Determine whether the series converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.
Solution 

video by Dr Chris Tisdell 

Final Answer 

The series converges absolutely by the direct comparison test and the absolute convergence theorem. 
close solution

Log in to rate this practice problem and to see it's current rating. 

\( \displaystyle{\sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }\)
Problem Statement 

For the series \( \displaystyle{\sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }\)
1. Determine whether the series converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.
Final Answer 

The series converges absolutely.
Problem Statement 

For the series \( \displaystyle{\sum_{n=1}^{\infty}{ \frac{n^3+n}{n^5+4} } }\)
1. Determine whether the series converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.
Solution 

video by PatrickJMT 

Final Answer 

The series converges absolutely. 
close solution

Log in to rate this practice problem and to see it's current rating. 

You CAN Ace Calculus
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
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
To bookmark this page and practice problems, log in to your account or set up a free account.
Single Variable Calculus 

MultiVariable Calculus 

Differential Equations 

Precalculus 

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.
 
free ideas to save on books 

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free. 
Practice Instructions
Unless otherwise instructed, for each of the following series:
1. Determine whether it converges or diverges.
2. Determine if a convergent series converges absolutely or conditionally.