Absolute Convergence

Measuring a Series Without Relying on Cancellation

The previous tutorial used alternating signs to locate a sum and estimate the error after finitely many terms. Those estimates depend on cancellation between successive terms. A different question is whether the terms would still have a finite total size if their signs were removed. This leads to absolute convergence, a property that can often be tested with positive-term series.

Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The series of magnitudes, $\sum_{n=1}^{\infty}|a_n|$, has nonnegative terms. We will call the original series absolutely convergent when this magnitude series converges. In this tutorial, the focus is on recognizing and estimating absolute convergence; the relationship between absolute convergence and convergence of the original signed series is taken up in the next tutorial.

Definition: The series $\sum_{n=1}^{\infty}a_n$ is absolutely convergent if the series $$ \sum_{n=1}^{\infty}|a_n| $$ converges. A series that converges but is not absolutely convergent is called conditionally convergent.

Absolute convergence is a statement about the sizes of the terms, not their signs. It is therefore natural to use the positive-series tools established earlier: comparison, the p-series criterion, and geometric bounds. For instance, if the magnitudes are eventually no larger than the terms of a convergent positive series, their sum is finite as well.

The Cauchy Criterion for Absolute Tails

A useful way to recognize absolute convergence is to ask whether every sufficiently late finite block has small total magnitude. This is the Cauchy Criterion for Series applied to the sequence $(|a_n|)$. The absolute values are essential: a signed block can be small because of cancellation even when its terms have large total magnitude.

Theorem (Cauchy Criterion for Absolute Convergence): The series $\sum_{n=1}^{\infty}a_n$ is absolutely convergent if and only if, for every $\varepsilon>0$, there is an integer $N$ such that for all integers $q\geq p\geq N$, $$ \sum_{n=p}^{q}|a_n|<\varepsilon. $$

Proof. Apply the Cauchy Criterion for Series to the series with terms $|a_n|$. Its partial sums are $A_m=\sum_{n=1}^{m}|a_n|$. For $q\geq p$, the difference between the partial sums immediately after and immediately before the block is $$ A_q-A_{p-1}=\sum_{n=p}^{q}|a_n|. $$ Thus the partial sums $(A_m)$ satisfy the Cauchy condition precisely when every such finite block of magnitudes is eventually smaller than $\varepsilon$. Since a sequence of real numbers converges if and only if it is Cauchy, this is equivalent to convergence of $\sum |a_n|$, which is the definition of absolute convergence. $\square$

This criterion supplies a practical diagnostic. To show that a series is absolutely convergent, one can bound the total magnitudes in every sufficiently late block. To show it is not absolutely convergent, it is enough to find some fixed positive amount of magnitude that keeps appearing in arbitrarily late blocks. The signed Cauchy criterion alone does not provide this information, since positive and negative terms may offset one another.

Comparison with a Positive Series

The most direct test is comparison. The terms $|a_n|$ are nonnegative, so if they are eventually bounded by the terms of a convergent positive series, their partial sums remain bounded. The finitely many terms before the comparison begins cannot affect convergence; this is consistent with the earlier result that finite changes preserve series convergence.

Theorem (Comparison Test for Absolute Convergence): Suppose $a_n$ is real and $b_n\geq0$ for every $n$. If there is an integer $N$ such that $$ |a_n|\leq b_n\qquad\text{for every }n\geq N $$ and $\sum_{n=1}^{\infty}b_n$ converges, then $\sum_{n=1}^{\infty}a_n$ is absolutely convergent.

Proof. Define $A_m=\sum_{n=1}^{m}|a_n|$. The sequence $(A_m)$ is nondecreasing because each added term $|a_{m+1}|$ is nonnegative. For $m\geq N$, the comparison gives $$ A_m =\sum_{n=1}^{N-1}|a_n|+\sum_{n=N}^{m}|a_n| \leq \sum_{n=1}^{N-1}|a_n|+\sum_{n=N}^{m}b_n \leq \sum_{n=1}^{N-1}|a_n|+\sum_{n=N}^{\infty}b_n. $$ The right-hand side is finite and independent of $m$, because the comparison series converges. The partial sums $A_m$ are therefore nondecreasing and bounded above. By the earlier theorem on bounded increasing partial sums, $(A_m)$ converges. Hence $\sum |a_n|$ converges, so $\sum a_n$ is absolutely convergent. $\square$

In practice, this theorem lets us discard the signs and compare magnitudes with a familiar benchmark. If a term has a complicated formula, it can be enough to find a simpler upper bound valid for all sufficiently large indices. The comparison need not hold at the start of the series.

Worked Examples

Worked Example: A Rational Term Compared with a p-Series

Consider $$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{3n+1}{n^3+4}. $$ For $n\geq1$, $3n+1\leq4n$, since $4n-(3n+1)=n-1\geq0$. Also $n^3+4\geq n^3>0$. Therefore $$ \left|(-1)^{n-1}\frac{3n+1}{n^3+4}\right| =\frac{3n+1}{n^3+4} \leq\frac{4n}{n^3} =\frac{4}{n^2}. $$ The comparison series $\sum 4/n^2$ converges by the p-Series Convergence Criterion, since $2>1$. The Comparison Test for Absolute Convergence now shows that the given series is absolutely convergent. The alternating signs are not needed for this conclusion.

Worked Example: Factorial Terms with a Geometric Tail Bound

Consider $$ \sum_{n=1}^{\infty}(-1)^n\frac{5^n}{n!}. $$ The magnitudes are $c_n=5^n/n!>0$. Their consecutive ratios are $$ \frac{c_{n+1}}{c_n} =\frac{5^{n+1}}{(n+1)!}\frac{n!}{5^n} =\frac{5}{n+1}. $$ For every $n\geq9$, $5/(n+1)\leq1/2$. Consequently $c_{n+1}\leq c_n/2$ for $n\geq9$, and induction gives $$ c_n\leq c_9\left(\frac12\right)^{n-9}\qquad(n\geq9). $$ The geometric series with these bounds converges because its ratio is $1/2<1$. By comparison, $\sum c_n$ converges, so the original series is absolutely convergent. The first eight terms cause no difficulty: a finite sum of magnitudes is finite.

Worked Example: Alternating Harmonic Terms Are Not Absolutely Summable

Consider the alternating harmonic series $$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}. $$ Its magnitudes form the harmonic series: $$ \sum_{n=1}^{\infty}\left|(-1)^{n-1}\frac{1}{n}\right| =\sum_{n=1}^{\infty}\frac{1}{n}. $$ This is the p-series with $p=1$, so it diverges by the p-Series Convergence Criterion. The alternating series itself converges by the Alternating Series Test: $1/n$ is nonincreasing, nonnegative, and tends to zero. Thus it converges but is not absolutely convergent; it is conditionally convergent. This example shows why convergence of a signed series alone does not settle whether the sum of its term magnitudes is finite.

What Absolute Convergence Gives for Finite Tails

Even before considering the sum of the original signed series, absolute convergence gives a useful bound for every finite block of its terms. The triangle inequality ensures that cancellation can only reduce the magnitude of a signed block; it cannot make that magnitude exceed the total size of the terms in the block.

Proposition (Absolute Tail Bound): Suppose $\sum_{n=1}^{\infty}|a_n|$ converges. For integers $q\geq p\geq1$, $$ \left|\sum_{n=p}^{q}a_n\right|\leq\sum_{n=p}^{q}|a_n|. $$ Moreover, for every $\varepsilon>0$, there is an $N$ such that the absolute value of every finite block $\sum_{n=p}^{q}a_n$, with $q\geq p\geq N$, is less than $\varepsilon$.

Proof. The finite triangle inequality gives $$ \left|\sum_{n=p}^{q}a_n\right|\leq\sum_{n=p}^{q}|a_n|, $$ which proves the first assertion. Since $\sum |a_n|$ converges, the Cauchy Criterion for Absolute Convergence supplies, for any $\varepsilon>0$, an $N$ such that $\sum_{n=p}^{q}|a_n|<\varepsilon$ whenever $q\geq p\geq N$. Combining this with the first inequality yields $$ \left|\sum_{n=p}^{q}a_n\right|<\varepsilon. $$ This proves the second assertion. $\square$

The proposition is useful when estimating errors or checking Cauchy conditions: one can bound a signed finite tail by a positive tail and avoid analyzing the pattern of signs. Absolute convergence therefore provides control that remains valid even if the signs are irregular.

A Common Pitfall: Cancellation Is Not Absolute Convergence

A small signed partial sum does not, by itself, indicate that the magnitudes have a small total. For example, a block containing a large positive term and a large negative term can sum to zero while the sum of their absolute values is large. The Cauchy Criterion for Absolute Convergence avoids this pitfall by testing the magnitude of every term in a late block, not just the signed total.

When testing a series, keep the two questions separate. First, ask whether $\sum |a_n|$ converges; comparison with a p-series or geometric series is often effective. If it does not, the original signed series may still converge through cancellation, as the alternating harmonic example demonstrates. The next tutorial will establish the general theorem relating absolute convergence to convergence of the original series.

Takeaway: Absolute convergence means convergence of the series of term magnitudes. Compare $|a_n|$ with a known convergent positive series to prove it, and use absolute tails to bound every sufficiently late finite block independently of cancellation.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What series must converge for $\sum a_n$ to be absolutely convergent?
Why does an eventual bound $|a_n|\leq b_n$, with $\sum b_n$ convergent and $b_n\geq0$, suffice to prove absolute convergence?
What condition on finite blocks of $|a_n|$ characterizes absolute convergence?
Why does convergence of the alternating harmonic series not imply absolute convergence?
How does the absolute tail bound control a signed finite block?

What You'll Learn