A series \sum a_n can converge for two very different reasons.
Either its terms are small enough that even their sizes
|a_n| add up to something finite — a robust, unconditional kind of
convergence — or the series only converges because positive and negative terms
cancel, a fragile arrangement that falls apart the moment you reshuffle it. The
distinction is the whole point of this page.
A series \sum a_n converges absolutely when the
series of absolute values converges:
\sum_{n=1}^{\infty} |a_n| \ \text{converges}.
It converges conditionally when \sum a_n
converges but \sum |a_n| does not. The headline theorem
is that the first kind is genuinely stronger: stripping the signs and still converging is
enough to guarantee the original series converges too.
Absolute convergence implies convergence
Suppose \sum |a_n| converges. We want to conclude that
\sum a_n converges. The trick is to bound the messy signed term by
something non-negative whose series we already control.
Step 1 — sandwich the shifted term. For every real
a_n we have -|a_n| \le a_n \le |a_n|.
Add |a_n| throughout to make the left edge zero:
0 \ \le \ a_n + |a_n| \ \le \ 2\,|a_n|.
Step 2 — name the non-negative series. Set
b_n = a_n + |a_n|. By Step 1 every b_n \ge 0,
and each is dominated by 2|a_n|.
Step 3 — compare. The series
\sum 2|a_n| = 2\sum |a_n| converges (a convergent series times a
constant). Since 0 \le b_n \le 2|a_n|, the
comparison test
forces \sum b_n to converge as well.
\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty}\big(a_n + |a_n|\big) \ \text{converges}.
Step 4 — undo the shift. Now solve for a_n:
a_n = b_n - |a_n|. Both \sum b_n and
\sum |a_n| converge, and the difference of two convergent series
converges:
\sum_{n=1}^{\infty} a_n \ = \ \sum_{n=1}^{\infty} b_n \ - \ \sum_{n=1}^{\infty} |a_n| \ \text{converges}.
Step 5 — read off the conclusion. We started only with the convergence of
\sum |a_n| and derived the convergence of
\sum a_n. So absolute convergence is the stronger property —
it comes for free with ordinary convergence on top.
Let \sum a_n be a series of real terms.
-
If \sum |a_n| converges, then
\sum a_n converges — absolute convergence implies
convergence.
-
The converse fails: \sum a_n may converge while
\sum |a_n| diverges. Such a series is
conditionally convergent.
-
The model example is the alternating harmonic series
\sum (-1)^{n+1}/n = \ln 2, which converges, whereas
\sum 1/n (the harmonic series) diverges to
\infty. So the alternating harmonic series is conditional.
Conditional convergence is not merely weaker; it is structurally unstable. Bernhard
Riemann proved a startling fact: if \sum a_n converges
conditionally, then for any target
L \in [-\infty, +\infty] you can reorder its terms so that the
rearranged series converges to L.
The mechanism is exactly the failure of absolute convergence. Split the terms into the
positives and the negatives. Because \sum|a_n| diverges while
\sum a_n converges, the positive part
\sum a_n^{+} and the negative part
\sum a_n^{-} must each diverge — there is an infinite
reservoir of positive sum and an infinite reservoir of negative sum. To hit a target
L, add just enough positive terms to climb above
L, then just enough negative terms to drop below it, and repeat;
since the individual terms shrink to 0, the overshoot shrinks too
and the partial sums close in on L.
The punchline: for a conditionally convergent series, the symbol "\sum a_n"
secretly depends on the order of summation. Absolute convergence is precisely the
condition that immunises a series against this — rearrange an absolutely convergent series
however you like and the sum never moves.