Source: Integral Calculus, University of Texas at Austin | MCQ Topic Quiz Notes
Difficulty: Intermediate | Prerequisites: comfort with sequences, basic series definitions, geometric series formula.
In integral calculus, you will frequently encounter infinite series and need to determine whether they converge or diverge. The convergence tests covered here are core tools for that job: the nth-term test, the limit comparison test, and direct comparison with geometric series. These sit between the introductory work on sequences and the later material on power series, Taylor series, and Maclaurin series. If you cannot reliably apply these tests, the power-series chapters will be rough going, because Taylor series convergence depends on them.
You should already be comfortable with limits, L'Hôpital's rule, and the formula for a convergent geometric series before working through these notes.
Convergence tests let you decide whether an infinite series adds up to a finite number or blows up to infinity. The nth-term test is a quick first check: if the terms do not shrink to zero, the series diverges. For everything else, comparison-based tests (limit comparison, direct comparison with a known geometric series) are your workhorses.
Convergence (of a series)
A series converges if the sequence of its partial sums approaches a finite limit. In simple terms, the total you get by adding more and more terms settles down to a specific number.
Divergence (of a series)
A series diverges if the partial sums do not approach a finite limit. Think of it as: the running total either grows without bound, oscillates, or otherwise refuses to settle.
Nth-Term Test (Test for Divergence)
If lim(n→∞) aₙ ≠ 0, the series Σaₙ diverges. In simple terms, if the individual terms are not shrinking to zero, there is no chance the series converges. Note: the converse is not true. Terms going to zero does not guarantee convergence.
Limit Comparison Test (LCT)
Given two series Σaₙ and Σbₙ with positive terms, compute L = lim(n→∞) aₙ/bₙ. If 0 < L < ∞ (a positive, finite number), both series either converge or both diverge. Think of it as: the two series behave the same way in the long run if their terms grow or shrink at the same rate.
Parent Function (Comparison Series)
The simpler, known series you compare against when using the limit comparison test or direct comparison test. You strip away lower-order terms from the original series to find it. In simple terms, this is the "simplified version" that you already know converges or diverges.
Geometric Series
A series of the form Σarⁿ (or equivalently Σrⁿ). It converges if |r| < 1 and diverges if |r| ≥ 1. This is the most common parent function used in comparison tests.
Harmonic Series
The series Σ(1/n), which diverges despite its terms approaching zero. This is the classic counterexample showing that the nth-term test is only a one-way test.
The nth-term test is the fastest check in your toolkit. It answers one question only: can this series be ruled out immediately?
If lim(n→∞) aₙ ≠ 0, the series Σaₙ diverges. Full stop.
If lim(n→∞) aₙ = 0, the test is inconclusive. The series might converge or might diverge. You need a different test.
This example comes directly from the source quiz material.
Step 1: Apply the nth-term test.
Compute lim(n→∞) sin(1/n).
As n → ∞, 1/n → 0, so sin(1/n) → sin(0) = 0.
The nth-term test is inconclusive here. It does not tell us the series converges; it only fails to prove divergence.
Step 2: Use the limit comparison test instead.
Compare Σ sin(1/n) with the harmonic series Σ(1/n), which is known to diverge.
Compute lim(n→∞) sin(1/n) / (1/n).
This is the classic limit: lim(x→0) sin(x)/x = 1.
Since L = 1 (positive and finite), the LCT says both series behave the same way.
The harmonic series diverges, so Σ sin(1/n) also diverges.
The key insight: sin(1/n) behaves like 1/n for large n. The sine function barely changes its input when the input is near zero, so the series inherits the harmonic series' divergence.
The limit comparison test is your go-to when you have a series that looks like a known series but has extra terms muddying the picture.
You have a mystery series Σaₙ.
You pick a comparison series Σbₙ whose convergence or divergence you already know.
Compute L = lim(n→∞) aₙ / bₙ.
If 0 < L < ∞, both series share the same fate: both converge or both diverge.
This example comes from the source quiz. The question asks which series to use for limit comparison.
Step 1: Identify the parent function.
Look at the dominant terms. As n grows large, 5ⁿ dwarfs n², so 5ⁿ − n² behaves like 5ⁿ.
The series Σ 4ⁿ / (5ⁿ − n²) therefore behaves like Σ 4ⁿ / 5ⁿ = Σ (4/5)ⁿ.
Step 2: Recognise the parent.
Σ (4/5)ⁿ is a geometric series with ratio r = 4/5.
Since |r| = 4/5 < 1, this geometric series converges.
Step 3: Apply the LCT.
With bₙ = (4/5)ⁿ as the comparison series, you would compute the limit of aₙ/bₙ, confirm it is positive and finite, and conclude the original series also converges.
The technique in one line: strip away the lower-order noise (the −n² in the denominator) to reveal the geometric backbone.
The hardest part of comparison tests is choosing the right series to compare against. Here is the process.
Look at the general term aₙ and identify which parts grow fastest as n → ∞.
Drop additive constants and lower-order terms from both numerator and denominator.
Simplify what remains into a recognisable form (geometric, p-series, or harmonic).
From the source quiz, this series is analysed for convergence.
Step 1: Find the parent function.
The dominant term in the denominator is 3ⁿ (the +n is negligible for large n).
So 2ⁿ / (3ⁿ + n) behaves like 2ⁿ / 3ⁿ = (2/3)ⁿ.
Step 2: Check the parent.
Σ (2/3)ⁿ is geometric with r = 2/3 < 1, so it converges.
Step 3: Confirm with the limit comparison test.
Compute lim(n→∞) [2ⁿ/(3ⁿ + n)] / [(2/3)ⁿ].
This simplifies to lim(n→∞) 3ⁿ/(3ⁿ + n) = 1.
Since L = 1 (positive, finite), the original series converges by the LCT.
Step 4: Alternatively, use direct comparison.
Note that 2ⁿ/(3ⁿ + n) < 2ⁿ/3ⁿ = (2/3)ⁿ for all n ≥ 1.
Since (2/3)ⁿ converges and our terms are smaller, the original series converges by direct comparison.
Terms with aⁿ/bⁿ patterns → geometric series Σ(a/b)ⁿ, converges when a/b < 1
Terms with 1/nᵖ patterns → p-series, converges when p > 1
Terms with 1/n pattern → harmonic series, diverges
Terms with sin(1/n) or similar → compare with 1/n (they behave alike near zero)
Nth-Term Test
If lim(n→∞) aₙ ≠ 0, then Σaₙ diverges.
Limit Comparison Test
Given Σaₙ and Σbₙ with aₙ, bₙ > 0:
L = lim(n→∞) aₙ / bₙ
If 0 < L < ∞, both series converge or both diverge.
Geometric Series
Σ(n=0 to ∞) rⁿ = 1/(1 − r) when |r| < 1; diverges when |r| ≥ 1.
Key Limit
lim(x→0) sin(x)/x = 1. This is the reason sin(1/n) and 1/n behave alike for large n.
Convergence tests matter wherever infinite sums appear in practice. Signal processing relies on Fourier series (infinite sums of sines and cosines), and engineers need to know those sums converge before trusting the output. In numerical computing, algorithms often approximate functions using truncated series, and convergence speed determines how many terms you need for a given accuracy. Even in finance, the present value of a perpetuity (an infinite stream of payments) is a geometric series, and the convergence condition |r| < 1 is what makes the valuation finite.
Students often think that if lim(n→∞) aₙ = 0, the series must converge. It does not. The harmonic series Σ(1/n) is the textbook counterexample: terms go to zero, but the series diverges. The nth-term test can only prove divergence, never convergence.
Students sometimes confuse the limit comparison test with the direct comparison test. The LCT uses a limit of the ratio aₙ/bₙ. The direct comparison test requires you to show aₙ ≤ bₙ (or aₙ ≥ bₙ) for all n beyond some point. They are related but not interchangeable in every situation.
When finding a parent function, students often forget to drop lower-order terms. The series Σ 4ⁿ/(5ⁿ − n²) should be compared with Σ(4/5)ⁿ, not with Σ 4ⁿ/5ⁿ − Σ 4ⁿ/n². You simplify the whole fraction, not split it into separate sums.
Students sometimes pick a parent function with the wrong convergence behaviour, making the comparison useless. If your mystery series looks like it converges, compare with a convergent series; if it looks like it diverges, compare with a divergent one.
⚠️ The nth-term test is a common first question on exams. You will be given a series and asked whether the nth-term test applies. Remember: it can only prove divergence. If the limit is zero, say "inconclusive" and move on.
⚠️ Expect at least one question asking you to identify the correct comparison series for the limit comparison test. Practice stripping away lower-order terms quickly.
⚠️ The limit lim(x→0) sin(x)/x = 1 appears frequently. Know it cold. It is the bridge between sin(1/n) and the harmonic series.
⚠️ Geometric series convergence (|r| < 1) is tested both on its own and as a parent function inside comparison problems. Be ready to spot the ratio r buried inside a more complex expression.
True or false: If lim(n→∞) aₙ = 0, the series Σaₙ converges.
False. The harmonic series is the counterexample.
Fill in the blank: The limit comparison test requires the limit L = lim(n→∞) aₙ/bₙ to be ______ and ______ for both series to share the same convergence behaviour.
Positive and finite (i.e. 0 < L < ∞).
True or false: Σ(3/4)ⁿ converges.
True. It is geometric with |r| = 3/4 < 1.
Fill in the blank: The parent function for Σ 2ⁿ/(3ⁿ + n) is Σ______.
(2/3)ⁿ.
True or false: lim(x→0) sin(x)/x = 0.
False. The limit equals 1.
Q: Does the series Σ sin(1/n) converge or diverge? Justify your answer.
A: It diverges. The nth-term test is inconclusive (lim sin(1/n) = 0), but by the limit comparison test with the harmonic series Σ(1/n), the limit of sin(1/n)/(1/n) = 1, and since the harmonic series diverges, so does Σ sin(1/n).
Q: Which series would you use with the limit comparison test to determine whether Σ 4ⁿ/(5ⁿ − n²) converges or diverges?
A: Use Σ(4/5)ⁿ. Stripping the −n² from the denominator leaves 4ⁿ/5ⁿ = (4/5)ⁿ. This is a convergent geometric series (r = 4/5 < 1), and the LCT confirms the original series converges.
Q: The series Σ 2ⁿ/(3ⁿ + n) from n = 1 to ∞. Which statement is true: it diverges by the nth-term test, or it converges by comparison?
A: It converges by comparison to Σ(2/3)ⁿ. The nth-term test does not apply here because lim(n→∞) 2ⁿ/(3ⁿ + n) = 0. The parent function is (2/3)ⁿ, a convergent geometric series.
Q: Give an example of a series whose terms approach zero but which nonetheless diverges.
A: The harmonic series Σ(1/n). The terms 1/n → 0, but the partial sums grow without bound.
Q: Explain why the nth-term test cannot prove convergence.
A: The test only checks a necessary condition (terms must go to zero for convergence to be possible). It is not a sufficient condition. The harmonic series satisfies the necessary condition but still diverges, which proves the test cannot work in the other direction.
These convergence tests feed directly into the study of power series, Taylor series, and Maclaurin series. When you write a Taylor series, you need to find its interval of convergence, which means applying ratio tests and comparison tests to determine where the series produces a finite answer. The limit comparison test in particular shows up when proving convergence of remainder terms in Taylor approximations.
The geometric series formula reappears constantly in later chapters. It is the simplest power series (a power series in r with all coefficients equal to 1), and understanding why it converges only for |r| < 1 is the template for understanding radius of convergence more broadly.
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