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Infinite Sequences and Series
Determine when infinitely many contributions produce a finite result.
What you’ll learn
- Distinguish sequence limits from series convergence.
- Choose and justify convergence tests.
- Build power-series representations and bound approximation errors.
Before you begin
A sequence is an ordered list of terms. A series adds terms and is studied through its partial sums. A necessary condition is not automatically sufficient: terms must approach zero for a series to converge, but that condition alone does not guarantee convergence.
Explain these starting ideas in your own words. Revisit them whenever a later step feels unclear.
A series is a limit of partial sums
A sequence lists terms an; a series adds them through partial sums Sn. The series converges if its partial sums approach a finite limit. Terms approaching zero are necessary for convergence but not sufficient: the harmonic series has terms tending to zero and still diverges.
A geometric series has a constant term ratio r and converges for |r|<1, with sum when a is its first term. If |r|≥1, a nonzero geometric series diverges. Check starting index so the first term is not mistaken for a coefficient from a different index.
PAUSE & TRY IT
What does a nonzero limit of series terms imply?
Reveal answer
The series diverges.
Positive-term tests compare long-run size
A p-series converges for p>1 and diverges for p≤1. The integral test applies when a matching function is positive, continuous, and decreasing eventually. It links convergence of a series with its corresponding improper integral, not equality of their values.
Direct comparison uses a valid inequality with nonnegative terms in the correct direction. Limit comparison uses a finite positive limit of term ratios to transfer convergence behavior. A bound below by a convergent series proves nothing about convergence, and an upper bound by a divergent series proves nothing about divergence.
Alternation can support conditional convergence
An alternating series converges under the standard test when term magnitudes decrease eventually to zero. If the absolute-value series also converges, convergence is absolute; if only the signed series converges, it is conditional. Apply the decreasing-magnitude and zero-limit checks explicitly.
For an alternating series satisfying the conditions, truncation error is at most the magnitude of the first omitted term. This bounds the approximation without requiring the exact sum. Do not apply that bound to an arbitrary oscillating or nondecreasing series.
PAUSE & TRY IT
What distinguishes absolute from conditional convergence?
Reveal answer
Absolute convergence also holds after taking absolute values of all terms; conditional convergence does not.
The ratio test has an inconclusive boundary
The ratio test examines the limit of |an+1/an|. A limit less than one gives absolute convergence; greater than one gives divergence; equal to one is inconclusive. Factorials and exponential factors often simplify well with this test.
An inconclusive result is not a divergence result. Choose another test suited to the remaining form. Before elaborate work, check whether terms fail to approach zero or whether the expression is a recognizable geometric or p-series.
Read figure values as text
sin x: -2: -0.9092974268256817; -1.9: -0.9463000876874145; -1.8: -0.9738476308781951; -1.7: -0.9916648104524686; -1.6: -0.9995736030415051; -1.5: -0.9974949866040544; -1.4: -0.9854497299884601; -1.3: -0.963558185417193; -1.2: -0.9320390859672263; -1.1: -0.8912073600614354; -1: -0.8414709848078965; -0.8999999999999999: -0.7833269096274833; -0.8: -0.7173560908995228; -0.7: -0.644217687237691; -0.6000000000000001: -0.5646424733950355; -0.5: -0.479425538604203; -0.3999999999999999: -0.3894183423086504; -0.30000000000000004: -0.2955202066613396; -0.19999999999999996: -0.19866933079506116; -0.10000000000000009: -0.09983341664682824; 0: 0; 0.10000000000000009: 0.09983341664682824; 0.20000000000000018: 0.19866933079506138; 0.2999999999999998: 0.2955202066613394; 0.3999999999999999: 0.3894183423086504; 0.5: 0.479425538604203; 0.6000000000000001: 0.5646424733950355; 0.7000000000000002: 0.6442176872376912; 0.7999999999999998: 0.7173560908995227; 0.8999999999999999: 0.7833269096274833; 1: 0.8414709848078965; 1.1: 0.8912073600614354; 1.2000000000000002: 0.9320390859672264; 1.2999999999999998: 0.963558185417193; 1.4: 0.9854497299884601; 1.5: 0.9974949866040544; 1.6: 0.9995736030415051; 1.7000000000000002: 0.9916648104524686; 1.7999999999999998: 0.9738476308781953; 1.9: 0.9463000876874145; 2: 0.9092974268256817 • Degree 1: -2: -2; 2: 2 • Degree 3: -2: -0.6666666666666667; -1.9: -0.7568333333333335; -1.8: -0.828; -1.7: -0.8811666666666668; -1.6: -0.9173333333333332; -1.5: -0.9375; -1.4: -0.9426666666666668; -1.3: -0.9338333333333333; -1.2: -0.9119999999999999; -1.1: -0.8781666666666667; -1: -0.8333333333333334; -0.8999999999999999: -0.7785; -0.8: -0.7146666666666667; -0.7: -0.6428333333333333; -0.6000000000000001: -0.5640000000000001; -0.5: -0.4791666666666667; -0.3999999999999999: -0.38933333333333325; -0.30000000000000004: -0.29550000000000004; -0.19999999999999996: -0.19866666666666663; -0.10000000000000009: -0.09983333333333343; 0: 0; 0.10000000000000009: 0.09983333333333343; 0.20000000000000018: 0.19866666666666685; 0.2999999999999998: 0.2954999999999998; 0.3999999999999999: 0.38933333333333325; 0.5: 0.4791666666666667; 0.6000000000000001: 0.5640000000000001; 0.7000000000000002: 0.6428333333333335; 0.7999999999999998: 0.7146666666666666; 0.8999999999999999: 0.7785; 1: 0.8333333333333334; 1.1: 0.8781666666666667; 1.2000000000000002: 0.912; 1.2999999999999998: 0.9338333333333333; 1.4: 0.9426666666666668; 1.5: 0.9375; 1.6: 0.9173333333333332; 1.7000000000000002: 0.8811666666666667; 1.7999999999999998: 0.8280000000000002; 1.9: 0.7568333333333335; 2: 0.6666666666666667 • Degree 5: -2: -0.9333333333333333; -1.9: -0.9631749166666668; -1.8: -0.985464; -1.7: -0.9994880833333334; -1.6: -1.0047146666666666; -1.5: -1.00078125; -1.4: -0.9874853333333334; -1.3: -0.9647744166666666; -1.2: -0.9327359999999999; -1.1: -0.8915875833333333; -1: -0.8416666666666667; -0.8999999999999999: -0.78342075; -0.8: -0.7173973333333333; -0.7: -0.6442339166666666; -0.6000000000000001: -0.564648; -0.5: -0.47942708333333334; -0.3999999999999999: -0.3894186666666666; -0.30000000000000004: -0.29552025000000004; -0.19999999999999996: -0.1986693333333333; -0.10000000000000009: -0.09983341666666676; 0: 0; 0.10000000000000009: 0.09983341666666676; 0.20000000000000018: 0.19866933333333353; 0.2999999999999998: 0.2955202499999998; 0.3999999999999999: 0.3894186666666666; 0.5: 0.47942708333333334; 0.6000000000000001: 0.564648; 0.7000000000000002: 0.6442339166666669; 0.7999999999999998: 0.7173973333333332; 0.8999999999999999: 0.78342075; 1: 0.8416666666666667; 1.1: 0.8915875833333333; 1.2000000000000002: 0.932736; 1.2999999999999998: 0.9647744166666666; 1.4: 0.9874853333333334; 1.5: 1.00078125; 1.6: 1.0047146666666666; 1.7000000000000002: 0.9994880833333334; 1.7999999999999998: 0.9854640000000001; 1.9: 0.9631749166666668; 2: 0.9333333333333333
Power series converge on an interval
A power series Σcn(x−a)n has a center a and a radius of convergence that may be zero, finite, or infinite. A ratio-test inequality often gives the open interval. Test each endpoint separately because one, both, or neither can converge.
Within the interval of convergence, power series can be differentiated or integrated term by term; the radius stays the same, but endpoint behavior may change. Substitution into known series gives new representations with transformed interval restrictions. State those restrictions with the representation.
Taylor coefficients encode derivatives at a center
The nth Taylor coefficient at a is f⁽ⁿ⁾!, so a Taylor polynomial matches successive derivatives at that center. A Maclaurin series is centered at zero. Common representations for ex, sin x, cos x, and provide useful building blocks.
An infinite Taylor series represents the function only where its remainder tends to zero; writing coefficients alone is not a universal proof of equality. Polynomial approximations are local, and more terms require an error argument if a specified accuracy is requested.
Read figure values as text
sin x: -3: -0.1411200080598672; -2.875: -0.26344599336342084; -2.75: -0.38166099205233167; -2.625: -0.4939202986100892; -2.5: -0.5984721441039565; -2.375: -0.6936850319532718; -2.25: -0.7780731968879212; -2.125: -0.850319789818452; -2: -0.9092974268256817; -1.875: -0.9540857816096938; -1.75: -0.9839859468739369; -1.625: -0.9985313405398316; -1.5: -0.9974949866040544; -1.375: -0.9808930570231557; -1.25: -0.9489846193555862; -1.125: -0.9022675940990952; -1: -0.8414709848078965; -0.875: -0.7675435022360271; -0.75: -0.6816387600233341; -0.625: -0.5850972729404622; -0.5: -0.479425538604203; -0.375: -0.36627252908604757; -0.25: -0.24740395925452294; -0.125: -0.12467473338522769; 0: 0; 0.125: 0.12467473338522769; 0.25: 0.24740395925452294; 0.375: 0.36627252908604757; 0.5: 0.479425538604203; 0.625: 0.5850972729404622; 0.75: 0.6816387600233341; 0.875: 0.7675435022360271; 1: 0.8414709848078965; 1.125: 0.9022675940990952; 1.25: 0.9489846193555862; 1.375: 0.9808930570231557; 1.5: 0.9974949866040544; 1.625: 0.9985313405398316; 1.75: 0.9839859468739369; 1.875: 0.9540857816096938; 2: 0.9092974268256817; 2.125: 0.850319789818452; 2.25: 0.7780731968879212; 2.375: 0.6936850319532718; 2.5: 0.5984721441039565; 2.625: 0.4939202986100892; 2.75: 0.38166099205233167; 2.875: 0.26344599336342084; 3: 0.1411200080598672 • x−x³/6: -3: 1.5; -2.875: 1.0856119791666665; -2.75: 0.7161458333333335; -2.625: 0.3896484375; -2.5: 0.10416666666666652; -2.375: -0.14225260416666652; -2.25: -0.3515625; -2.125: -0.5257161458333333; -2: -0.6666666666666667; -1.875: -0.7763671875; -1.75: -0.8567708333333334; -1.625: -0.9098307291666666; -1.5: -0.9375; -1.375: -0.9417317708333333; -1.25: -0.9244791666666667; -1.125: -0.8876953125; -1: -0.8333333333333334; -0.875: -0.7633463541666666; -0.75: -0.6796875; -0.625: -0.5843098958333334; -0.5: -0.4791666666666667; -0.375: -0.3662109375; -0.25: -0.24739583333333334; -0.125: -0.12467447916666667; 0: 0; 0.125: 0.12467447916666667; 0.25: 0.24739583333333334; 0.375: 0.3662109375; 0.5: 0.4791666666666667; 0.625: 0.5843098958333334; 0.75: 0.6796875; 0.875: 0.7633463541666666; 1: 0.8333333333333334; 1.125: 0.8876953125; 1.25: 0.9244791666666667; 1.375: 0.9417317708333333; 1.5: 0.9375; 1.625: 0.9098307291666666; 1.75: 0.8567708333333334; 1.875: 0.7763671875; 2: 0.6666666666666667; 2.125: 0.5257161458333333; 2.25: 0.3515625; 2.375: 0.14225260416666652; 2.5: -0.10416666666666652; 2.625: -0.3896484375; 2.75: -0.7161458333333335; 2.875: -1.0856119791666665; 3: -1.5 • x−x³/6+x⁵/120: -3: -0.5249999999999999; -2.875: -0.5512346903483074; -2.75: -0.5944905598958332; -2.625: -0.648992156982422; -2.5: -0.7096354166666669; -2.375: -0.7719571431477863; -2.25: -0.8321044921875; -2.125: -0.886804453531901; -2: -0.9333333333333333; -1.875: -0.9694862365722656; -1.75: -0.9935465494791667; -1.625: -1.0042554219563802; -1.5: -1.00078125; -1.375: -0.9826891581217447; -1.25: -0.9499104817708334; -1.125: -0.9027122497558594; -1: -0.8416666666666667; -0.875: -0.767620595296224; -0.75: -0.6816650390625; -0.625: -0.5851046244303386; -0.5: -0.47942708333333334; -0.375: -0.3662727355957031; -0.25: -0.24740397135416667; -0.125: -0.12467473347981771; 0: 0; 0.125: 0.12467473347981771; 0.25: 0.24740397135416667; 0.375: 0.3662727355957031; 0.5: 0.47942708333333334; 0.625: 0.5851046244303386; 0.75: 0.6816650390625; 0.875: 0.767620595296224; 1: 0.8416666666666667; 1.125: 0.9027122497558594; 1.25: 0.9499104817708334; 1.375: 0.9826891581217447; 1.5: 1.00078125; 1.625: 1.0042554219563802; 1.75: 0.9935465494791667; 1.875: 0.9694862365722656; 2: 0.9333333333333333; 2.125: 0.886804453531901; 2.25: 0.8321044921875; 2.375: 0.7719571431477863; 2.5: 0.7096354166666669; 2.625: 0.648992156982422; 2.75: 0.5944905598958332; 2.875: 0.5512346903483074; 3: 0.5249999999999999
Error bounds make approximations accountable
A Lagrange remainder bound uses a bound M on the magnitude of the (n+1)st derivative between the center and target: |Rn| ≤ M|x−a|!. M must bound the derivative over the entire relevant interval, not just at the center.
For an appropriate alternating representation, the first-omitted-term bound may be simpler. State which theorem and conditions justify the bound. An error bound is an upper bound, not necessarily the exact error or its sign.
PAUSE & TRY IT
Where must the derivative bound M hold for a Lagrange remainder?
Reveal answer
Throughout the interval between the expansion center and the target input.
Choose a convergence test from the structure
Start with the term test: if terms fail to approach zero, the series diverges. A geometric series has a constant ratio and converges when the ratio’s magnitude is below one. A p-series ∑ᵖ converges only for p>1. For a positive-term series resembling one of these, comparison or limit comparison can transfer a known conclusion when the hypotheses hold.
Factorials and exponentials often suggest the ratio test. If the limiting absolute ratio is below one, the series converges absolutely; above one, it diverges; equal to one, the test is inconclusive. Inconclusive does not mean divergent. State the comparison series or calculated ratio and the conclusion rather than naming a test without applying it.
PAUSE & TRY IT
What does a ratio-test limit of 1 tell you?
Reveal answer
Only that this test is inconclusive; use another test.
Distinguish absolute, conditional, and estimated convergence
An alternating series can converge when term magnitudes decrease to zero. Check both conditions, at least eventually, and use the alternating-series error bound only when its hypotheses apply. The error after N terms is no larger than the magnitude of the first omitted term. This is a bound on the unknown error, not necessarily its exact value.
Absolute convergence means the series of absolute values converges. Conditional convergence means the original series converges while its absolute-value series diverges. A convergent alternating harmonic series is a standard structural example of the latter distinction. Use the appropriate tests on both the original and absolute-value series to justify the classification.
PAUSE & TRY IT
Why are power-series endpoints checked individually?
Reveal answer
Substitution can produce different series at the two endpoints, with different convergence behavior.
Power series need a radius and endpoint tests
A power series centered at a has terms involving (x−a)ⁿ. The ratio test typically gives an open interval of absolute convergence. Test the two endpoints separately by substituting each endpoint into the original series. One endpoint can converge while the other diverges. Reporting only the radius leaves the interval incomplete.
Taylor coefficients are f⁽ⁿ⁾!, so the polynomial matches successive derivatives at its center. A Maclaurin series is centered at zero. Substitution, differentiation, and integration of known series can construct new ones inside the interval of convergence, but endpoint behavior must be rechecked when needed. A Taylor polynomial is an approximation, and a Lagrange or alternating-series bound can quantify its error under the relevant assumptions.
Separate a sequence from its series
A sequence lists terms an; a series sums them. If a series converges, its terms must approach zero, but that condition is not sufficient. The harmonic series has terms approaching zero and still diverges. Apply the nth-term test to detect divergence, not to prove convergence.
A geometric series has constant ratio r and converges when |r|<1. Its sum depends on the first included term. A p-series ∑ᵖ converges for p>1 and diverges for p≤1. Recognizing these reference forms helps with comparison tests.
A finite number of initial terms does not change whether an infinite series converges, but it changes its sum. Keep starting indices straight when calculating an exact geometric sum or building a polynomial approximation.
Choose a convergence test that matches the structure
Positive-term comparisons require valid inequalities or a suitable limit comparison. Compare dominant behavior for rational expressions, but state the reference series and why it converges or diverges. An integral test needs positive, continuous, decreasing behavior eventually for the associated function.
Ratio and root tests are useful for factorials or exponential powers. A limit less than one gives absolute convergence; greater than one gives divergence; equal to one is inconclusive. Inconclusive means choose another test, not that the series diverges.
An alternating-series test needs term magnitudes approaching zero and eventually decreasing. Absolute convergence means the series of magnitudes converges; conditional convergence means the original converges while the magnitude series diverges. Show both parts when classifying conditional convergence.
PAUSE & TRY IT
The ratio-test limit is 1. What can you conclude?
Reveal answer
The test is inconclusive. Another argument is required; both convergent and divergent series can have ratio-test limit 1.
Build power series and test endpoints separately
A power series is centered at a specific value. The ratio test often gives an open interval of convergence; substitute each endpoint into the original series and test it separately. One endpoint can converge while the other diverges. The radius alone does not describe endpoint behavior.
Known series can be transformed by substitution, multiplication, differentiation, or integration within their valid intervals. Track changes to coefficients, powers, and indices. Integrating can change endpoint convergence even though the radius remains the same, so endpoints need renewed attention.
Taylor coefficients are derivatives at the center divided by factorials. A Taylor polynomial approximates locally; a Taylor series represents the function where the remainder tends to zero. Matching several derivatives does not justify claiming exact equality everywhere.
Use an error bound with its hypotheses
For a qualifying alternating series, the remainder magnitude is at most the first omitted term’s magnitude. Choose enough terms to make that bound smaller than the required tolerance. The actual error may be smaller, but the bound is what guarantees the requested accuracy.
A Lagrange remainder bound uses a maximum bound on the appropriate derivative over the interval between center and evaluation point. State that bound and the factorial denominator. Using the derivative only at the center may fail to bound it throughout the interval.
When reporting an approximation, include the polynomial value and a justified error statement. A calculator decimal alone is not evidence of a guaranteed tolerance. If the requested accuracy is strict, check whether equality at the bound is sufficient or whether the bound must be strictly smaller.
Build a test-selection argument rather than naming tests at random
First check whether terms approach zero. Then inspect recognizable structure: geometric ratio, p-series power, factorials, alternating signs, or comparison with a simpler positive series. State the test’s conditions and the result that actually follows. A list of test names does not justify convergence.
For ∑, compare with . The ratio of terms approaches 1, a finite positive constant, so limit comparison with the convergent p-series establishes convergence. The dominant-power observation guides the choice, while the explicit ratio supplies the justification.
For an alternating series, classify absolute convergence separately. ∑(−1)ⁿ/n converges by the alternating-series test but its magnitude series is harmonic and diverges. This is conditional convergence. Omitting the magnitude-series check leaves the classification incomplete.
PAUSE & TRY IT
Why is “the terms go to zero” insufficient to prove convergence?
Reveal answer
The harmonic series is a counterexample: its terms approach zero but its partial sums grow without bound.
Choose the polynomial degree from the error requirement
For a Taylor approximation, begin with the expansion center and the required evaluation point. Write enough terms to see the pattern, then determine which remainder bound is available. In an alternating series with decreasing term magnitudes, the first omitted term can establish accuracy.
For sin(0.2), the approximation 0.2− omits a next term of magnitude ≈0.00000267. The alternating-series remainder is no larger than that under the usual conditions. This supplies a reason for the number of terms, rather than choosing a degree because it looks sophisticated.
With a Lagrange bound, bound the relevant derivative on the whole interval from the center to the evaluation point. If the bound is too large, increase the degree or choose a better justified derivative bound. Keep the distinction between a guaranteed upper bound and the actual unknown error.
FROM IDEA TO APPLICATION
Worked examples
Terms going to zero are not enough
Classify Σ from n=1 to infinity of and explain why the nth-term test cannot prove convergence.
Reveal worked solution
- The terms approach zero, satisfying only a necessary condition.
- It is a p-series with p=1.
It diverges. A zero term limit does not establish convergence.
Check both endpoints
Find the convergence interval of Σ from n=1 to infinity of xn/n.
Reveal worked solution
- The ratio test gives |x|<1.
- At x=1, the harmonic series diverges.
- At x=−1, the alternating harmonic series converges.
[−1,1).
Bound a sine approximation
Approximate sin(0.2) using 0.2 − and bound the error with the alternating series.
Reveal worked solution
- The approximation is 0.1986666667.
- The first omitted magnitude is ≈ 0.0000026667.
- The sine terms decrease in magnitude here and approach zero.
The absolute error is at most about 2.67 × 10-6.
Find and test both endpoints
Find the interval of convergence of ∑ from n=1 to infinity of (x−2)ⁿ/(n3ⁿ).
Reveal worked solution
- The absolute ratio limit is |x−2|/3, so convergence is guaranteed for |x−2|<3.
- At x=−1, the series becomes ∑(−1)ⁿ/n and converges by the alternating-series test.
- At x=5, it becomes ∑ and diverges.
The interval is [−1,5), with radius 3.
Use an error bound to choose a stopping point
An alternating series has decreasing term magnitudes . How many terms guarantee error at most 0.001?
Reveal worked solution
- The alternating-series error is at most 3.
- Require 3≤0.001, so N+1≥10.
Nine terms are sufficient under the alternating-series hypotheses.
Determine endpoints, not just radius
Find the interval of convergence of ∑ from n=1 to infinity of (x−2)ⁿ/n.
Reveal worked solution
- The ratio test gives |x−2|<1, hence 1<x<3.
- At x=1, the series is ∑(−1)ⁿ/n, which converges conditionally.
- At x=3, it is the divergent harmonic series.
The interval is [1,3); the radius is 1.
A limit comparison
Determine convergence of ∑ from n=1 to infinity of .
Reveal worked solution
- Compare with bn=.
- an/bn=→1.
- The p-series with p=2 converges.
The given positive series converges by limit comparison.
A justified sine approximation
Approximate sin(0.2) using x− and bound error.
Reveal worked solution
- Value = 0.2−≈0.1986667.
- Next term magnitude is ≈2.67×10-6.
- Alternating decreasing terms justify using that remainder bound.
sin(0.2)≈0.1986667 with error at most about 0.00000267.
MAKE THE DISTINCTION
Common mistakes, clearer reasoning
The trapRatio-test limit 1 means divergence.
The better explanationIt is inconclusive; use another test.
The trapThe radius determines endpoint inclusion.
The better explanationEach endpoint must be tested separately.
RETRIEVE BEFORE YOU REVEAL
Practice checkpoints
Revisit the quick checks from this guide without looking back. Explain why, then reveal the answer.
1. What distinguishes absolute from conditional convergence?
Reveal answer
Absolute convergence also holds after taking absolute values of all terms; conditional convergence does not.
2. What does a nonzero limit of series terms imply?
Reveal answer
The series diverges.
3. Where must the derivative bound M hold for a Lagrange remainder?
Reveal answer
Throughout the interval between the expansion center and the target input.
4. What does a ratio-test limit of 1 tell you?
Reveal answer
Only that this test is inconclusive; use another test.
5. Why are power-series endpoints checked individually?
Reveal answer
Substitution can produce different series at the two endpoints, with different convergence behavior.
6. The ratio-test limit is 1. What can you conclude?
Reveal answer
The test is inconclusive. Another argument is required; both convergent and divergent series can have ratio-test limit 1.
7. Why is “the terms go to zero” insufficient to prove convergence?
Reveal answer
The harmonic series is a counterexample: its terms approach zero but its partial sums grow without bound.
Key language
- Partial sum
- The sum of finitely many initial series terms.
- Absolute convergence
- Convergence of the series of absolute values.
- Radius of convergence
- The distance from the center defining the power series’ open convergence region.
- Remainder
- The difference between a function or series sum and a finite approximation.
Series extend finite polynomial approximations into infinite representations with explicit convergence and error control.