
is said to converge absolutely if

converges. If the series does converge but does not converge absolutely, then it is said to converge conditionally.
Proof: Let ε > 0 be given. Then by assumption and Theorem 5.1.7, there exists n_{ε} such that whenever q ≥ p ≥ n_{ε},

Therefore, from the triangle inequality,

By Theorem 5.1.7, ∑ _{k=m}^{∞}a_{k} converges. ■
In fact, the above theorem is really another version of the completeness axiom. Thus its validity implies completeness. You might try to show this.
One of the interesting things about absolutely convergent series is that you can “add them up” in any order and you will always get the same thing. This is the meaning of the following theorem. Of course there is no problem when you are dealing with finite sums thanks to the commutative law of addition. However, when you have infinite sums strange and wonderful things can happen because these involve a limit.
Proof: From absolute convergence, there exists M such that

Since θ is one to one and onto, there exists N ≥ M such that

This is because the partial sums of the above series are each dominated by a partial sum for ∑ _{k=M+1}^{∞}

Hence


Since ε is arbitrary, this shows the two series are equal as claimed. ■
So what happens when series converge only conditionally?
Example 5.1.11 Consider the series ∑ _{k=1}^{∞}
First of all consider why it converges. Notice that if S_{n} denotes the n^{th} partial sum, then
Theorem 5.1.12 (comparison test) Suppose
Proof: Consider the first claim. From the assumption, there exists n^{∗} such that n^{∗} > max
For n > 1,

Now
A convenient way to implement the comparison test is to use the limit comparison test. This is considered next.
Theorem 5.1.14 Let a_{n},b_{n} > 0 and suppose for all n large enough,

Then ∑ a_{n} and ∑ b_{n} converge or diverge together.
Proof:Let n^{∗} be such that n ≥ n^{∗}, then

and so for all such n,

and so the conclusion follows from the comparison test. ■
The following corollary follows right away from the definition of the limit.
Corollary 5.1.15 Let a_{n},b_{n} > 0 and suppose

Then ∑ a_{n} and ∑ b_{n} converge or diverge together.
Example 5.1.16 Determine the convergence of ∑ _{k=1}^{∞}
This series converges by the limit comparison test above. Compare with the series of Example 5.1.13.
To really exploit this limit comparison test, it is desirable to get lots of examples of series, some which converge and some which do not. The tool for obtaining these examples here will be the following wonderful theorem known as the Cauchy condensation test.
Theorem 5.1.17 Let a_{n} ≥ 0 and suppose the terms of the sequence

converge or diverge together.
Proof:This follows from the inequality of the following claim.
Claim:

Proof of the Claim: Note the claim is true for n = 1. Suppose the claim is true for n. Then, since 2^{n+1} − 2^{n} = 2^{n}, and the terms, a_{n}, are decreasing,


Example 5.1.18 Determine the convergence of ∑ _{k=1}^{∞}
Let a_{n} =

converge or diverge together. If p > 1, the last series above is a geometric series having common ratio less than 1 and so it converges. If p ≤ 1, it is still a geometric series but in this case the common ratio is either 1 or greater than 1 so the series diverges. It follows that the p series converges if p > 1 and diverges if p ≤ 1. In particular, ∑ _{n=1}^{∞}n^{−1} diverges while ∑ _{n=1}^{∞}n^{−2} converges.
Example 5.1.19 Determine the convergence of ∑ _{k=1}^{∞}
Use the limit comparison test.

and so this series diverges with ∑ _{k=1}^{∞}
Sometimes it is good to be able to say a series does not converge. The n^{th} term test gives such a condition which is sufficient for this. It is really a corollary of Theorem 5.1.7.
Proof:Apply Theorem 5.1.7 to conclude that

It is very important to observe that this theorem goes only in one direction. That is, you cannot conclude the series converges if lim_{n→∞}a_{n} = 0. If this happens, you don’t know anything from this information. Recall lim_{n→∞}n^{−1} = 0 but ∑ _{n=1}^{∞}n^{−1} diverges. The following picture is descriptive of the situation.