Convergence tests and absolute convergence

Small terms leave the main question open

The terms 1/n tend to zero, but their totals grow without bound. Changing the signs to 1−1/2+1/3−1/4+⋯ produces a convergent series. Changing the magnitudes to 1/n² gives another convergent series, even with every sign positive. We will justify both convergence claims and explain what the different arguments allow us to conclude.

A convergence test is an implication with hypotheses. It may prove existence of a finite sum without giving its value. Some tests also give a bound for the omitted tail. Distinguish these tasks before calculating: does the ordered sum exist, does the sum of absolute values exist, and how accurate is a chosen partial sum?

Preparation: use the preceding lessons on sequences, partial sums, geometric series and finite-tail identities. The integral test also uses improper integrals and elementary antiderivatives. Follow the comparison and alternating examples first if you need a route without integration; return to the optional rearrangement argument after the main tests.

Begin with the necessary condition aₙ→0. If it fails, Σaₙ diverges. If it holds, continue: it does not prove convergence. Removing finitely many defined initial terms preserves convergence, so a test may be applied to a sufficiently late tail. Every denominator and logarithm still needs its stated domain.

Absolute convergence controls cancellation

The series Σaₙ converges absolutely when Σ|aₙ| converges. It converges conditionally when Σaₙ converges but Σ|aₙ| diverges. The word conditionally is not a synonym for “not yet tested.” Both parts of its definition must be established.

Absolute convergence implies convergence. Define aₙ⁺=max(aₙ,0) and aₙ⁻=max(−aₙ,0). They are nonnegative, aₙ=aₙ⁺−aₙ⁻ and |aₙ|=aₙ⁺+aₙ⁻. If the absolute partial sums have a finite bound, the partial sums of both nonnegative parts have that bound too. The monotone bounded theorem makes both parts converge; their difference is the original series.

The finite triangle inequality also passes to the limit: |Σaₙ|≤Σ|aₙ|. Applying the same argument to the tail shows that the absolute value of a signed remainder is at most the sum of the omitted absolute values. A bound on the absolute tail therefore bounds the signed error, even when the exact sum is unknown.

Compare in the direction that supplies a bound

Suppose 0≤aₙ≤bₙ for every sufficiently large n. If Σbₙ converges, then Σaₙ converges: every finite a-tail total is bounded by the corresponding b-tail total. Conversely, if Σaₙ diverges, then Σbₙ diverges. Nonnegative partial sums can only diverge by being unbounded, and the larger totals cannot remain bounded.

For example 0<1/(3ⁿ+1)<1/3ⁿ for n≥1. The geometric series Σ3⁻ⁿ converges, so Σ1/(3ⁿ+1) converges. Each added term is smaller than a corresponding term of a known convergent series.

For a divergence comparison, √n≤n for n≥1, so 1/(n+√n)≥1/(2n). Half the harmonic series still has unbounded totals. Therefore Σ1/(n+√n) diverges.

The opposite directions do not decide anything: being below a divergent series or above a convergent one supplies no useful bound. The inequality 1/n²≤1/n, by itself, does not establish convergence of the smaller series. We need a separate convergence argument for it. For signed aₙ, compare |aₙ| with a nonnegative benchmark when proving absolute convergence.

A decreasing curve bounds discrete additions

Let f be continuous, nonnegative and nonincreasing on [m,∞), where m is a positive integer, and suppose aₙ=f(n) for n≥m. On each interval [n,n+1], f(n+1)≤f(x)≤f(n). Integrating over this unit interval turns these heights into lower and upper bounds on areas.

For integers M≥m, the integral from m to M+1 is at most the finite sum f(m)+⋯+f(M). That sum is at most f(m) plus the integral from m to M. If the improper integral from m to infinity converges, the nonnegative partial sums have a finite upper bound and converge. If it diverges to +∞, the lower integral bound makes the totals unbounded. This is the integral test.

The sum and integral generally have different values. The proof compares them; it does not identify them. Eventual decrease is enough because the finite prefix is separate. A few sampled decreasing values do not establish the required inequality for every real x in the tail.

For f(x)=1/x² on [1,∞), the improper integral equals 1. Thus Σₙ₌₁∞1/n² converges and its sum is at most 2 by the upper bound. We have proved convergence without evaluating the exact series sum.

Move the first rectangle when estimating the omitted tail

Under the same hypotheses, and when the series converges, define I(c) as the improper integral of f from the real lower endpoint c to infinity. The nonnegative remainder satisfies I(N+1)≤R_N≤I(N) for every integer N≥m. The upper integral begins one index earlier than the lower integral.

To check those endpoints, for n≥N+1 the integral over [n,n+1] is at most f(n), which is at most the integral over [n−1,n]. Add only through a finite M first. The integral from N+1 to M+1 bounds the finite tail below; the integral from N to M bounds it above. Letting M grow gives exactly I(N+1)≤R_N≤I(N).

For aₙ=1/n², this becomes 1/(N+1)≤R_N≤1/N. Choosing N=101 guarantees R_N≤1/101<1/100. This is a sufficient accuracy choice from the stated weak bound; it is not a claim that 101 is the least possible stopping index.

Weak inequalities are deliberate for a general nonincreasing function, which may have flat portions. Extra strictness needs an argument. The diagram displays only a finite group of tail rectangles; the inequalities above justify the entire infinite remainder.

Benchmark series have precise thresholds

A p-series is Σₙ₌₁∞1/nᵖ for a fixed real p. It converges exactly when p>1. For p>0 the function x⁻ᵖ is positive, continuous and decreasing, so the integral test applies. When p≠1, its antiderivative is x¹⁻ᵖ/(1−p); the improper integral converges for p>1 and diverges for 0<p<1. At p=1 the logarithm grows without bound. For p≤0, the terms themselves do not approach zero.

The conclusion depends on the exponent, not on whether the first few terms look small. Multiplying by a fixed positive constant or starting later does not change convergence. For p>1, the same integration gives the tail bounds (N+1)¹⁻ᵖ/(p−1)≤R_N≤N¹⁻ᵖ/(p−1).

A useful refinement begins at n=2: Σ1/[n(log n)ᑫ], where log is the natural logarithm and q is fixed and real. Substituting u=log x into the improper integral gives an integral of u⁻ᑫ from log 2 to infinity. It converges exactly when q>1. The original function is positive and continuous for x>1. Its logarithmic derivative is −(1+q/log x)/x, so it is decreasing sufficiently far out for every fixed q.

Consequently Σₙ₌₂∞1/[n(log n)²] converges, while Σₙ₌₂∞1/[n log n] diverges. Both are smaller than 1/n eventually. This supplies concrete examples of why “smaller than harmonic” is inconclusive.

A positive finite ratio transfers convergence

Suppose aₙ≥0 and bₙ>0 eventually, and aₙ/bₙ→L with 0<L<∞. For sufficiently large n, L/2<aₙ/bₙ<3L/2, so (L/2)bₙ≤aₙ≤(3L/2)bₙ. These two comparisons show that Σaₙ and Σbₙ either both converge or both diverge.

For aₙ=(3n+2)/(n³+7), compare with bₙ=1/n². The ratio is (3n³+2n²)/(n³+7), which tends to 3. The benchmark p-series converges, so Σ(3n+2)/(n³+7) converges. All terms and denominators are positive for n≥1.

For this example the bounds already hold for n≥2: 3/2≤aₙ/bₙ≤9/2. Multiplying by the positive n³+7 reduces the first inequality to 3n³+4n²≥21 and the second to 4n²≤3n³+63. Both hold for n≥2. The first term need not satisfy the lower bound; it belongs to the finite prefix.

There are useful one-way variants. If aₙ/bₙ→0 and Σbₙ converges, then eventually aₙ≤bₙ, so Σaₙ converges. If aₙ/bₙ→+∞ and Σbₙ diverges, then eventually aₙ≥bₙ, so Σaₙ diverges. An unbounded ratio alone is not the same as a ratio tending to +∞: isolated large values do not provide an eventual lower bound.

If the benchmark diverges and the ratio tends to zero, there is no conclusion from this test. Both 1/n² and 1/[n log n] have ratio tending to zero relative to 1/n, but their series have different convergence behaviour. Keep the direction and the limit value together.

A ratio below one eventually supplies a geometric tail

Suppose aₙ≠0 for every sufficiently large n and L=lim |aₙ₊₁/aₙ| exists, allowing +∞. If L<1, the series converges absolutely. Choose a fixed q with L<q<1. The limit guarantees |aₙ₊₁|≤q|aₙ| for all sufficiently large n; repeated use bounds the tail by a convergent geometric series.

If L>1, including L=+∞, choose 1<r<L. Eventually the magnitudes grow by at least the factor r, starting from a nonzero term, so they cannot tend to zero and the series diverges. If L=1, the ratio test is inconclusive. If the eventual denominators vanish, the test in this form is unavailable; that is not a divergence conclusion.

For aₙ=2ⁿ/n!, where n!=1·2·⋯·n, the ratio is 2/(n+1)→0. Hence Σₙ₌₁∞2ⁿ/n! converges absolutely. The first two terms happen to be equal; an early exception to decrease has no effect on the argument about a whole tail.

The proof can give an error bound. If |aₙ₊₁/aₙ|≤q<1 for all n≥N, then |R_N|≤|a_N₊₁|/(1−q). For the factorial example and N=9, every ratio from that point is at most 1/5. The first omitted term is a₁₀=2¹⁰/10!=4/14175, so R₉≤(4/14175)/(4/5)=1/2835<1/1000. No exact value of the infinite sum is needed.

An nth power may make the root test simpler

Let L=lim |aₙ|¹ᐟⁿ exist, again allowing +∞. If L<1, choose L<q<1; eventually |aₙ|≤qⁿ, proving absolute convergence by comparison. If L>1, then eventually |aₙ| is bounded below by rⁿ for some r>1, violating the zero-term condition. At L=1 the root test gives no conclusion.

For aₙ=[2n/(5n+1)]ⁿ, the nth root is 2n/(5n+1)→2/5, so the series converges absolutely. In fact the root is already below 2/5 at every positive integer, which also gives a direct geometric comparison.

Zero terms cause no problem for nth roots. More generally, finding a fixed q<1 with |aₙ|¹ᐟⁿ≤q for every sufficiently large n is sufficient even when the root sequence has no limit. For example, let aₙ=3⁻ⁿ at odd indices and 0 at even indices. Its roots alternate between 1/3 and 0, yet |aₙ|≤3⁻ⁿ proves absolute convergence. Failure of one limit-based test does not settle the series question.

One is a boundary value, not a verdict

For aₙ=1/n, the magnitude ratio is n/(n+1)→1 and the series diverges. For aₙ=1/n², the ratio is [n/(n+1)]²→1 and the series converges. This pair alone proves that ratio limit 1 cannot decide the answer.

Their root limits are also 1. Indeed, n⁻ᵖᐟⁿ=exp[−p(log n)/n]→1 for each fixed p, since (log n)/n→0. The p-series threshold, not the root limit, distinguishes them.

An inconclusive result is useful information about the argument: change the test. Simplify rational powers for comparison, use an integral for a suitable positive decreasing function, or check a signed pattern directly. Do not turn “this test does not decide” into “the series diverges.”

Two nested sets of partial sums can prove convergence

Consider Σₙ₌₁∞(−1)ⁿ⁺¹bₙ, where bₙ≥0, bₙ is nonincreasing and bₙ→0. The alternating series test proves convergence. The same conclusion holds when these conditions and the alternating pattern begin only after a finite initial segment.

For the stated starting sign, even partial sums are nondecreasing because S₂ₖ₊₂−S₂ₖ=b₂ₖ₊₁−b₂ₖ₊₂≥0. Odd partial sums are nonincreasing because S₂ₖ₊₃−S₂ₖ₊₁=−b₂ₖ₊₂+b₂ₖ₊₃≤0. Every even partial sum is at most every odd partial sum: advancing either parity keeps it between the earlier endpoints. Both sets are bounded, so both converge.

The gap S₂ₖ₊₁−S₂ₖ=b₂ₖ₊₁ tends to zero, forcing the two limits to agree. The whole sequence of partial sums then converges, because every sufficiently late index is either even or odd and both tails approach that same S.

The limit lies in the closed interval between S_N and S_N₊₁. Its width is b_N₊₁, giving |R_N|≤b_N₊₁. A nonzero remainder points in the direction of the next term. The interval statement handles equality and zero remainders too.

For bₙ=1/n, the hypotheses hold and the alternating harmonic series converges. Its absolute series is harmonic and diverges, so the convergence is conditional. At N=20 the alternating error is at most 1/21. This bound concerns the ordered alternating sum, not the unsigned harmonic total.

Check the decreasing-magnitude hypothesis

Alternating signs and terms tending to zero are insufficient by themselves. Define a₂ₖ₋₁=2/k and a₂ₖ=−1/k for k≥1. Its terms tend to zero, but S₂ₘ=Σₖ₌₁ᵐ(2/k−1/k)=Σₖ₌₁ᵐ1/k. These even partial sums are unbounded, so the original series diverges.

The magnitude sequence fails to be eventually nonincreasing: for k>1, the next odd magnitude 2/(k+1) exceeds the preceding even magnitude 1/k. The larger positive contributions leave a harmonic amount behind after each pair.

This counterexample shows why the alternating test needs its hypotheses. A different series whose magnitudes are not decreasing may still converge by another argument. Likewise the next-term error estimate is a consequence of the alternating theorem, not a general rule for arbitrary signed or positive series.

Optional: why absolute convergence protects the order

A rearrangement uses every original term exactly once in a new order: mathematically it is specified by a bijection of the positive integers. For an absolutely convergent series, every rearrangement converges absolutely to the same sum.

Here is the tail argument. Given ε>0, choose an original cutoff M with the absolute tail below ε/2. The signed remainder after M is then below ε/2 in magnitude too. Eventually a rearranged partial sum contains every one of the finitely many original indices 1 through M. Its extra terms are some finite selection from the absolute tail, whose total magnitude is below ε/2. Its difference from the original sum is therefore below ε by the triangle inequality. The same reasoning applies at every later rearranged cutoff and to the series of absolute values.

Conditional convergence is different. Its nonnegative positive and negative parts each have unbounded totals; if one had a finite total, convergence of their difference would force the other to have a finite total as well, contradicting divergence of the absolute series.

To see the consequence, pick a target T. Take unused positive terms in their original order until the total is above T, then unused negative terms until it is below T, and repeat. Both one-sign totals are unbounded, so every crossing is possible and switching continues forever. Every nonzero term is eventually used. Zero terms, if present, can be inserted without changing totals.

Each crossing misses T by at most the magnitude of the last term used. Those magnitudes tend to zero, since an original convergent series has terms tending to zero and the rearrangement eventually exhausts every finite set of large terms. Between crossings the totals move monotonically between the nearby crossing values. Thus all the rearranged partial sums approach T. This does not change the original ordered sum; it constructs a different order, explaining why rearrangement requires care.

Finish with the hypotheses and an accuracy statement

A useful solution names a test, verifies its domain and tail conditions, states the resulting convergence type, and gives a remainder estimate when approximation is required. A numerical plot can suggest a comparison or show an interval shrinking; it cannot certify every later term.

For rational algebraic terms, first try a geometric or p-series comparison, or a positive finite limit ratio. For an eventually decreasing nonnegative function with a manageable antiderivative, try the integral test. Factorials often simplify in a ratio; nth powers often simplify under an nth root. For signs, test absolute convergence and then check whether a valid alternating argument adds information.

Use an error estimate with the same verified conditions: an absolute-tail comparison, the two integral bounds, a geometric ratio envelope or the alternating next-term bound. These bounds answer different questions and need not be the smallest possible bounds.

The following explorer compares actual finite data with the proved bounds. Predict which quantity is controlled before revealing it. In the next lesson, the same tests will determine where a power series converges; boundary points will need their own arguments.

Worked examples

Try it yourself

1. Does aₙ→0 alone prove convergence of Σaₙ?

   Show the answer and reasoning

   No.

   Both harmonic and inverse-square terms tend to zero, but their series have different behaviour.

2. Test Σₙ₌₁∞1/(3ⁿ+1).

   Show the answer and reasoning

   It converges.

   Its nonnegative terms are below 3⁻ⁿ, whose geometric series converges.

3. Does 1/n²≤1/n prove convergence of Σ1/n² by direct comparison with harmonic?

   Show the answer and reasoning

   No.

   An upper bound by a divergent series is inconclusive. The integral or p-series argument supplies the missing convergence proof.

4. For which real p does Σₙ₌₁∞1/nᵖ converge?

   Show the answer and reasoning

   Exactly p>1.

   For p>0 use the improper integral; for p≤0 the terms fail to approach zero.

5. Compare Σₙ₌₂∞1/[n log n] with Σₙ₌₂∞1/[n(log n)²].

   Show the answer and reasoning

   The first diverges and the second converges.

   The substitution u=log x turns their integrals into p-integrals with exponents 1 and 2.

6. If aₙ/bₙ→0 and Σbₙ diverges, what follows about Σaₙ?

   Show the answer and reasoning

   Nothing from that limit-comparison fact alone.

   Relative to 1/n, both 1/n² and 1/[n log n] have ratio tending to zero, but one series converges and the other diverges.

7. A ratio test gives L=1. Should the series be marked divergent?

   Show the answer and reasoning

   No; the result is inconclusive.

   The series of 1/n and of 1/n² both have ratio limit 1.

8. Terms equal 3⁻ⁿ at odd indices and 0 at even indices. Can the ordinary ratio formula be applied eventually?

   Show the answer and reasoning

   No, but the series converges absolutely by comparison.

   There are zero denominators arbitrarily late. Nevertheless |aₙ|≤3⁻ⁿ controls the whole tail.

9. For aₙ=[2n/(5n+1)]ⁿ, what root limit decides convergence?

   Show the answer and reasoning

   2/5<1.

   Take the nth root before the limit; the changing exponent disappears.

10. Is an alternating sign pattern with magnitudes tending to zero enough for the alternating test?

    Show the answer and reasoning

    No; eventual nonincrease of the magnitudes is also needed.

    The terms 2/k and −1/k in successive pairs tend to zero, yet each pair contributes 1/k.

11. Use |R_N|≤1/(N+1) to guarantee error strictly below 1/100. Give a sufficient N.

    Show the answer and reasoning

    N=100 is sufficient.

    Then |R₁₀₀|≤1/101<1/100. At N=99 this particular weak bound only gives ≤1/100 and does not certify strictness.

12. Which property guarantees that every rearrangement has the same sum?

    Show the answer and reasoning

    Absolute convergence.

    A finite absolute tail controls terms omitted or added by a sufficiently late rearranged cutoff. Conditional convergence alone does not protect the sum under arbitrary reordering.

What to carry forward

• A necessary zero-term limit is a first check, not a convergence proof.
• Absolute convergence implies convergence and bounds signed tail errors by absolute tails.
• Comparison needs eventual nonnegative inequalities in the correct direction; positive finite limit ratios transfer convergence.
• The integral test compares a suitable nonincreasing nonnegative continuous function with discrete partial sums; sum and integral are generally unequal.
• A p-series converges exactly for p>1, and logarithmic refinements need their own threshold.
• Ratio or root limits below 1 prove absolute convergence; limits above 1 prove divergence; 1 is inconclusive.
• The alternating next-term bound needs eventual nonincreasing magnitudes tending to zero, together with the stated sign pattern.
• Absolute convergence preserves the sum under every rearrangement; conditional convergence does not.