Source: Integral Calculus, University of Texas at Austin
Tags: series convergence, alternating series test, divergence test, limit comparison test, p-series, integral calculus
Series convergence is about deciding whether an infinite sum settles to a finite value or blows up. The main tools for this homework are the Alternating Series Test (AST), the Divergence Test, and the Limit Comparison Test. Knowing which test to reach for, and checking its conditions carefully, is most of the battle.
Difficulty: Intermediate
Prerequisites: familiarity with basic sequences and series (partial sums, geometric series), comparison tests, and p-series. You should be comfortable with limits and basic algebraic manipulation of series terms.
This material sits in the second half of a standard Integral Calculus (Calculus II) course, after you have covered sequences and basic series. The central question is always the same: does this series converge or diverge? The tests you apply here (alternating series, divergence, limit comparison, and the interplay between absolute and conditional convergence) form the toolkit you will use for the rest of the course, including power series and Taylor series.
If you are behind, make sure you are solid on p-series (when does 1/n^p converge?), the behaviour of 1/sqrt(n), and what it means for a sequence of terms to approach zero. Without those, the reasoning below will not stick.
Convergent series
A series whose partial sums approach a finite limit as n goes to infinity. In simple terms, the running total settles down to a number rather than growing forever.
Divergent series
A series that does not converge. The partial sums either grow without bound, oscillate, or otherwise fail to approach a single finite value. Think of it as a sum that never "lands."
Alternating series
A series whose terms switch sign: positive, negative, positive, negative, and so on. The general form is the sum of (-1)^(n-1) * b_n, where b_n > 0. In simple terms, the terms take turns adding and subtracting.
Alternating Series Test (AST) / Leibniz Test
An alternating series converges if (1) b_n is eventually decreasing and (2) the limit of b_n as n approaches infinity is 0. This is one of the gentlest convergence tests: it only requires the terms to shrink to zero, not quickly.
Divergence Test (nth-Term Test)
If the limit of a_n as n approaches infinity is not zero, the series diverges. This test can only prove divergence, never convergence. Think of it as a quick screening: if the terms do not even approach zero, there is no hope.
Limit Comparison Test
Given two series with positive terms, if the limit of a_n / b_n as n approaches infinity equals a finite positive number L > 0, then both series converge or both diverge. In simple terms, if two series "behave the same way" for large n, they share the same fate.
p-series
The series sum of 1/n^p. Converges when p > 1, diverges when p <= 1. This is the benchmark series you compare against in many problems.
Conditional convergence
A series that converges, but does not converge absolutely (the series of absolute values diverges). Alternating series that pass the AST but whose absolute-value version diverges are the classic example.
Absolute convergence
A series where the sum of |a_n| converges. Absolute convergence implies ordinary convergence, but not the reverse. Think of it as a stronger form of convergence: even without the sign cancellation, the series still converges.
Many of the homework problems hand you four or five series and ask which one converges. The strategy:
First, check whether the series is truly alternating. The factor (-1)^(n-1) or (-1)^(n+1) flips sign each term. But (-1)^(2n) = 1 for all n, so it does not alternate at all. Similarly, (-1)^3 is just -1, a constant, so that series is not alternating either.
If the series is not alternating, treat it as a positive-term series. A series like the sum of 1/(3 + sqrt(n)) behaves like 1/sqrt(n) for large n, which is a divergent p-series (p = 1/2).
If the series is genuinely alternating, apply the AST: check that the magnitude of the terms decreases and approaches zero.
For example, the sum of (-1)^(n-1) / (2.5 + sqrt(n)) is alternating, with b_n = 1/(2.5 + sqrt(n)). This is decreasing and approaches zero, so the series converges by the AST.
By contrast, the sum of (-1)^(n-1) * (5 + sqrt(n)) / (3 + sqrt(n)) has terms whose magnitude approaches 1 (not zero), so it fails the AST and diverges by the Divergence Test.
The Divergence Test is your first line of defence. Before applying any other test, ask: does the limit of a_n equal zero? If not, the series diverges, full stop.
Consider the series sum of (2n / (n^2 + 4)) * cos(2npi). Since cos(2npi) = 1 for all integers n, this simplifies to sum of 2n / (n^2 + 4). For large n, this behaves like 2/n, and the limit of 2n/(n^2 + 4) is 0. So the Divergence Test is inconclusive here. You need to compare with the harmonic series: since 2n/(n^2 + 4) is comparable to 1/n (which diverges), the series diverges by the Limit Comparison Test.
This test is your workhorse for positive-term series that look like a "messy" version of a known series. The idea:
Identify the dominant behaviour of a_n for large n.
Pick a comparison series b_n that captures that behaviour (usually a p-series).
Compute the limit of a_n / b_n. If the limit is a finite positive number, both series share the same convergence behaviour.
For the series sum of (4 + k + k^4) / sqrt(3 + k^2 + k^10), the dominant terms for large k are k^4 in the numerator and sqrt(k^10) = k^5 in the denominator, giving behaviour like k^4 / k^5 = 1/k. Compare with the harmonic series 1/k. The limit of [(4 + k + k^4) / sqrt(3 + k^2 + k^10)] / (1/k) is finite and positive, so the series diverges (since the harmonic series diverges).
Some problems ask for which values of a parameter a series converges. For the series sum from n=2 of (-1)^(n-1) * (ln n)^p / (2n):
The factor (-1)^(n-1) makes it alternating.
The magnitude is (ln n)^p / (2n).
For any real p, (ln n)^p / (2n) approaches 0 as n approaches infinity (exponentials and polynomials always beat logarithms).
The terms are eventually decreasing for any real p.
So the series converges by the AST for all real p. The answer is (-infinity, infinity).
Alternating Series Test (Leibniz Test)
The series sum of (-1)^(n-1) * b_n converges if:
b_n >= b_(n+1) for all n beyond some index (terms eventually decrease)
lim (n -> infinity) b_n = 0
Divergence Test
If lim (n -> infinity) a_n is not equal to 0, then the series sum of a_n diverges.
Limit Comparison Test
For series with positive terms a_n and b_n, if lim (n -> infinity) a_n / b_n = L, where 0 < L < infinity, then sum a_n and sum b_n either both converge or both diverge.
p-series reference
The series sum of 1/n^p converges if and only if p > 1.
Students often think (-1)^(2n) alternates sign. It does not. Since 2n is always even, (-1)^(2n) = 1 for every n. A series with this factor is not alternating and cannot be handled by the AST.
Students sometimes believe that if the terms approach zero, the series must converge. This is false. The harmonic series sum of 1/n has terms approaching zero but diverges. The terms approaching zero is necessary but not sufficient.
A common error is applying the AST without checking that b_n is decreasing. A series whose terms alternate in sign but do not decrease in magnitude can still diverge.
Students confuse the Divergence Test with a convergence test. If lim a_n = 0, the Divergence Test says nothing. It cannot confirm convergence.
⚠️ "Which series converges?" questions test whether you can spot fake alternating factors like (-1)^(2n) or (-1)^3. Read the exponent carefully before applying the AST.
⚠️ The Limit Comparison Test is the standard approach for rational-looking series involving roots and polynomials. Exams love pairing a messy series with a clean p-series.
⚠️ For "find all values of p" questions, remember that logarithmic factors never affect convergence of the alternating version, only the rate of decrease. The AST cares about the limit being zero and the terms decreasing, and ln(n)^p / (2n) satisfies both for all real p.
⚠️ cos(2npi) = 1 for all integer n. This appears as a trap: it looks like it oscillates, but it does not.
True or False: The series sum of (-1)^(2n) / n is alternating.
False. (-1)^(2n) = 1 for all n, so the series is just sum of 1/n.
True or False: If lim a_n = 0, the series sum of a_n converges.
False. The harmonic series is the classic counterexample.
Fill in the blank: The Limit Comparison Test requires both series to have ______ terms.
Positive.
True or False: cos(2npi) = 1 for every integer n.
True.
Fill in the blank: The p-series sum of 1/n^p converges when p > ______.
Q: You are given the series sum of (-1)^(n-1) / (2.5 + sqrt(n)). Does it converge?
A: Yes. It is alternating with b_n = 1/(2.5 + sqrt(n)), which is decreasing and approaches 0. Converges by the AST.
Q: Does the series sum of (2n / (n^2 + 4)) * cos(2npi) converge or diverge?
A: cos(2npi) = 1 for all n, so the series is sum of 2n/(n^2 + 4). For large n this behaves like 2/n. By the Limit Comparison Test with the harmonic series, it diverges.
Q: For which values of p does sum of (-1)^(n-1) * (ln n)^p / (2n) converge (starting from n=2)?
A: For all real p, i.e. (-infinity, infinity). The magnitude (ln n)^p / (2n) approaches 0 and is eventually decreasing for any p.
Q: The series sum of (4 + k + k^4) / sqrt(3 + k^2 + k^10). Convergent or divergent?
A: For large k, the terms behave like k^4 / k^5 = 1/k. The harmonic series 1/k diverges, so by the Limit Comparison Test, this series diverges.
Q: Which of (-1)^(2n) * 3 / (1 + sqrt(n)) and (-1)^(n-1) * (5 + sqrt(n)) / (3 + sqrt(n)) is convergent?
A: Neither. The first is not alternating ((-1)^(2n) = 1), so it behaves like 3/sqrt(n), which diverges. The second has terms whose magnitude approaches 1, so it diverges by the Divergence Test.
This material connects directly to absolute and conditional convergence (covered in the next set of notes), where you determine not just whether a series converges, but how strongly. The Ratio Test and Root Test, also covered separately, offer alternative tools when comparison tests are awkward.
Series convergence underpins power series and Taylor series later in the course: the interval of convergence of a power series is found using the Ratio or Root Test, and behaviour at endpoints often comes back to the tests covered here.
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