A Fourier series is an infinite sum. Every honest question this course asks about one is really a question about series convergence first. Does it actually add up to the function it's built from? Does it add up to anything at all? None of the machinery for answering "does this converge" is new. What's new is how often the answer will matter for the rest of this course to make sense.
A sequence is just a numbered list, one number per lap: a_1, a_2, a_3, \dots. A series is what you get by running a scoreboard total of that list, adding one more term every time n goes up by one. Think of a partial sum as a runner's total distance after n laps. Suppose each lap adds less new distance than the one before, and the shrinkage happens fast enough. Then the runner's total distance settles near one number, no matter how many more laps they run. That number is what "the series converges" means. Now suppose instead every lap adds close to the same amount every time. The total distance climbs forever with no settling point. That's divergence.
Absolute versus conditional convergence is the difference between a runner who only ever moves forward and one who alternates forward and backward steps. Strip away direction and look only at each step's actual length. If that forward-only version already settles at a finite total distance, the run is absolutely convergent. That's the strongest kind, since it doesn't even need the back-and-forth to help. Now suppose the forward-only version would run off to infinity on its own. The runner is actually alternating forward and backward steps, though, and the net effect still settles at one spot. That's conditional convergence. The cancellation between forward and backward steps is doing all the work, and it's a far more fragile kind of settling than the absolute case.
A sequence \{a_n\} is an infinite, ordered list of numbers a_1, a_2, a_3, \dots, one term for every positive integer n. It converges to L if a_n gets and stays arbitrarily close to L as n \to \infty. Formally, for every \varepsilon > 0 there's an index N past which every term satisfies |a_n - L| < \varepsilon. A series \sum a_n is what you get by adding a sequence's terms together, one at a time, forever; it converges when its own sequence of partial sums S_n = a_1 + \cdots + a_n converges to some finite limit. That limit is the series' sum. A geometric series \sum_{n=0}^{\infty} ar^n converges exactly when |r| < 1, to \frac{a}{1-r}. It's the one series family whose convergence and exact sum are both known in closed form, which is why it's the benchmark every other test is compared against.
A series can converge two different ways. It's absolutely convergent if \sum |a_n| itself converges. That's the strong case, where even discarding every sign still leaves a finite sum. It's conditionally convergent if \sum a_n converges but \sum |a_n| diverges. There the signs are doing real work, canceling enough that the alternating sum settles even though the sum of sizes alone runs away. Absolute convergence is the stronger guarantee: it implies ordinary convergence, but not the other way around.
Choosing which test to reach for is its own skill, separate from running any one test correctly. Before computing anything, name the series' shape. A factorial or a variable exponent in every term points at the Ratio Test. An outer (\cdot)^n wrapping the whole term points at the Root Test. A rational function that looks like a p-series for large n points at Direct or Limit Comparison. Alternating signs point at the Alternating Series Test. The shape decides the tool. Running the Ratio Test on \sum \frac{1}{n^2} works, technically, but a comparison to the known p-series answers it in one line.
Theorem
Absolute Convergence Implies Convergence
Recalled from Calculus II — Integral Calculus & Infinite Series · Absolute and Conditional Convergence — first taught there.
If \displaystyle\sum |a_n| converges, then \displaystyle\sum a_n converges too.
For every n, 0 \le a_n + |a_n| \le 2|a_n| (this holds whether a_n is positive, negative, or zero: if a_n\ge 0 the middle expression is 2a_n; if a_n<0 it's 0). Since \sum|a_n| converges, so does \sum 2|a_n|, and the Direct Comparison Test then gives that \sum (a_n+|a_n|) converges too. Writing a_n = (a_n+|a_n|) - |a_n| expresses the original series as a difference of two convergent series, and a difference of convergent series converges (term-by-term subtraction of two convergent partial sums converges to the difference of their limits).
What a_n \to L actually requires
Shrink \varepsilon. Watch the widget search for the smallest N after which every later term stays inside the band.
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A geometric partial sum climbing toward its exact limit
Slide n up. Watch the staircase level off at the dashed line instead of climbing forever.
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Worked example
Summing a geometric series exactly
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1
Evaluate \displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n.
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2
This is geometric with first term a = \frac{1}{3} (the n=1 term) and ratio r = \frac{1}{3}. Since |r| = \frac{1}{3} < 1, it converges.
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3\sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^n = \frac{a}{1-r} = \frac{1/3}{1-1/3} = \frac{1/3}{2/3} = \boxed{\frac{1}{2}}.
Sanity check: the first three terms alone, \frac13+\frac19+\frac1{27} \approx 0.481, are already closing in on 0.5.
Worked example
Absolute or only conditional?
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1
Determine whether \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n} (the alternating harmonic series) converges absolutely, conditionally, or not at all.
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2
Strip the signs: \sum\left|\frac{(-1)^n}{n}\right| = \sum\frac{1}{n}, the harmonic series, which diverges. So absolute convergence is out.
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3
The signed series itself still converges. Its terms shrink to 0 and alternate, and that alone is enough to force the partial sums to settle. Chapter 4's Dirichlet-type conditions make this precise for a Fourier series' own convergence.
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4
\boxed{\text{Conditionally convergent}}. A genuinely weaker guarantee than the geometric series above, since it depends on the alternating signs for cancellation rather than on the terms' sizes alone.
Worked example
A rational series recognized by its shape
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1
Determine whether \displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2+1} converges.
Common trap. This isn't a p-series itself (n^2+1, not n^2), so reaching for the p-series test directly doesn't apply. But for large n, the +1 barely matters, and that's the actual signal to compare against \frac{1}{n^2}.
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2
For every n \ge 1, 0 \le \frac{1}{n^2+1} \le \frac{1}{n^2}, and \sum \frac{1}{n^2} is a p-series with p=2>1, which converges.
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3
\boxed{\text{Converges}}, by Direct Comparison. The series is squeezed underneath a known convergent one.
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Name the tool before reaching for it
Answer the branch questions and land on the one test that actually fits this series.
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Where this goes next
Every infinite sum in this course
Chapter 4 asks, for every Fourier series this course writes down, whether it converges pointwise, uniformly, or not at all at a given point. Chapter 5's Parseval's identity is itself a statement about a series of squared coefficients converging to a finite energy. Neither question is answerable without the vocabulary built here. Both need what "converges" means for a sequence of partial sums. Both need the difference between a sum that settles because its terms are genuinely small and one that settles only because its signs cancel.
Check your understanding (counts toward your progress)
Classify \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}.
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Check the series of absolute values first: \sum\frac{1}{n^2} is a p-series with p=2>1, which already converges on its own.
The terms shrink to 0, and stripping the signs still leaves a convergent p-series — this doesn't diverge.
Why
\sum\left|\frac{(-1)^n}{n^2}\right| = \sum\frac1{n^2} converges (p=2>1), so the original series converges absolutely — the alternating signs aren't even needed for convergence here.
Strip the alternating sign and test \sum\frac{1}{n^2} on its own first.
Evaluate \displaystyle\sum_{n=1}^{\infty}\left(\frac14\right)^n.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Geometric with a=\frac14, r=\frac14: \frac{a}{1-r} = \frac{1/4}{3/4} = \frac13.
Correct answer
Also accepted:
Why
Which test is the natural first choice for \displaystyle\sum_{n=1}^{\infty}\frac{n!}{2^n}?
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There's no alternating sign here — every term is positive.
A factorial isn't shaped like a rational function comparable to \frac{1}{n^p}.
The n! in the numerator means the ratio between consecutive terms isn't constant, so this isn't geometric.
Why
A factorial in the terms is the classic signal for the Ratio Test, since \frac{a_{n+1}}{a_n} collapses the factorial down to a simplified ratio.
Look at what's in the terms — a factorial, an alternating sign, a rational function, or a constant ratio — each shape points to one test.
Practice: Convergence, a sum, and which test to reach for (4 problems · doesn’t affect your progress)
Does the sequence a_n = \dfrac{1}{n^2} converge? If so, to what limit?
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\frac{1}{n^2} shrinks toward 0 as n grows — it never settles near 1.
The terms get and stay arbitrarily close to a single finite number as n\to\infty, which is exactly what convergence means.
Why
As n\to\infty, \frac{1}{n^2}\to 0, and the terms stay arbitrarily close to 0 from then on — the sequence converges to 0.
Yes, to 0
Evaluate \displaystyle\sum_{n=0}^{\infty}(0.6)^n.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Geometric with first term a=1 (the n=0 term) and ratio r=0.6: \frac{a}{1-r}=\frac{1}{0.4}=\frac{5}{2}.
Correct answer
Also accepted:
Why
The sum starts at n=0, so the first term a is (0.6)^0=1, not 0.6 itself.
Classify \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt n}.
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Strip the signs first: \sum\frac{1}{\sqrt n} is a p-series with p=\frac12\le 1, which diverges — so absolute convergence is ruled out.
The terms shrink to 0 and alternate, which is enough to force the signed series itself to settle, even though the un-signed version runs away.
Why
\sum\left|\frac{(-1)^n}{\sqrt n}\right|=\sum\frac{1}{\sqrt n} diverges (p=\frac12\le1), but the alternating signed series still converges — conditionally convergent.
Conditionally convergent
For the series \displaystyle\sum_{n=1}^{\infty}\frac{3^n}{n!}, which convergence test should you reach for first?
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Every term here is positive — there's no alternating sign to exploit.
A factorial in the denominator isn't shaped like a rational function comparable to \frac1{n^p}.
The n! means consecutive terms don't share a constant ratio, so this isn't geometric.
Why
A factorial is the classic signal for the Ratio Test — \frac{a_{n+1}}{a_n} collapses the factorial into a simplified ratio that's easy to take a limit of.
Ratio Test