Difficulty: Intermediate | Prerequisites: Chapter 9 study notes, limits at infinity (Section 2.6), L'Hôpital's Rule (Section 4.5), improper integrals.
This is Part 1 of the Chapter 10 material. It covers the foundational ideas: what sequences and series are, how to determine whether they converge, and the first set of convergence tests (Integral Test, p-series, Direct Comparison, Limit Comparison). These tools form the backbone for every later convergence test. If you are behind, make sure you are comfortable with limits at infinity and improper integrals before starting.
A sequence is an ordered list of numbers; a series is the sum of a sequence's terms. The key question is always "does it converge or diverge?" Geometric series have a clean formula. For everything else, you compare, integrate, or take limits. The tests in this set handle series with positive terms.
Sequence
An ordered list of numbers {a₁, a₂, a₃, ...} defined by a formula for the nth term a_n. Think of it as a function whose input is the positive integers.
Convergent sequence
A sequence {a_n} for which lim(n→∞) a_n = L, where L is a finite number.
Divergent sequence
A sequence whose limit does not exist or is infinite.
Infinite series
The sum of all terms of an infinite sequence: Σ a_n = a₁ + a₂ + a₃ + ...
Partial sum (S_n)
The sum of the first n terms: S_n = a₁ + a₂ + ... + a_n. A series converges if the sequence of partial sums {S_n} converges. In simple terms, keep adding terms and see whether the running total settles down to a fixed number.
Geometric series
A series of the form Σ ar^(n−1), where a is the first term and r is the common ratio. Converges to a/(1–r) when |r| < 1; diverges otherwise.
Harmonic series
The series Σ 1/n = 1 + 1/2 + 1/3 + ... It diverges, even though its terms approach zero.
p-series
The series Σ 1/n^p. Converges when p > 1; diverges when p ≤ 1. Think of it as a family of series indexed by the exponent p, with p = 1 (the harmonic series) as the dividing line.
Integral Test
If f(x) is continuous, positive, and decreasing on [k, ∞) with f(n) = a_n, then the series and the improper integral ∫ f(x) dx either both converge or both diverge.
Direct Comparison Test
Compare a_n term by term against a known series b_n. If a_n ≤ b_n and Σ b_n converges, then Σ a_n converges. If a_n ≥ b_n and Σ b_n diverges, then Σ a_n diverges.
Limit Comparison Test
If lim(n→∞) a_n/b_n = c, where 0 < c < ∞, then Σ a_n and Σ b_n either both converge or both diverge.
A sequence is an ordered list of numbers generated by a formula for the nth term a_n. Notation: {a_n} or {a_n} from n = 1 to ∞. The input is always a positive integer; the output is the corresponding term.
A sequence can be thought of as a function whose domain is the positive integers. Its graph is a set of isolated points, not a continuous curve.
If lim(n→∞) a_n = L for some finite number L, the sequence is convergent. Otherwise it is divergent.
Examples:
a_n = n/(n+1)². Since deg(numerator) < deg(denominator), the limit is 0. Convergent.
a_n = n²/(n+2). Since deg(numerator) > deg(denominator), the limit is ∞. Divergent.
a_n = cos(nπ/2). The terms cycle through 0, −1, 0, 1, 0, −1, ... The limit does not exist. Divergent.
For a_n = p(n)/q(n) where p and q are polynomials:
deg(p) < deg(q): limit is 0.
deg(p) = deg(q): limit is the ratio of leading coefficients.
deg(p) > deg(q): limit is ±∞ (divergent).
Example: a_n = (3 + 5n²)/(n + n²). Leading coefficients are 5 and 1, so the limit is 5.
Squeeze Theorem. If a_n ≤ c_n ≤ b_n for all n > N, and lim a_n = lim b_n = L, then lim c_n = L.
Absolute value rule. If lim |a_n| = 0, then lim a_n = 0. Useful for alternating sequences like (−1)^n · n/(n²+1): the absolute value n/(n²+1) → 0, so the sequence converges to 0.
Geometric sequence {r^n}. Converges to 0 if |r| < 1. Converges to 1 if r = 1. Diverges for all other values of r.
Continuity and limits. If lim a_n = L and f is continuous at L, then lim f(a_n) = f(L). In other words, you can bring the limit inside a continuous function.
Bounded monotone sequences. Every bounded and increasing sequence converges. Every bounded and decreasing sequence converges. (You do not need to find the limit to conclude convergence.)
This is an indeterminate form of type 1^∞. To evaluate it, take the natural log, rewrite the product as a quotient, apply L'Hôpital's Rule, and exponentiate:
Let y = lim(n→∞) (1 + 1/n)^n.
ln y = lim n · ln(1 + 1/n) = lim ln(1 + 1/n) / (1/n), which is a 0/0 form.
Apply L'Hôpital's Rule: the limit evaluates to 1.
So ln y = 1, which gives y = e.
More generally: lim(n→∞) (1 + m/n)^n = e^m.
Given an infinite sequence {a_n}, the infinite series is the sum Σ a_n = a₁ + a₂ + a₃ + ... The nth partial sum is S_n = a₁ + a₂ + ... + a_n.
The series converges if the sequence of partial sums {S_n} converges to a finite limit s. We then write Σ a_n = s. If {S_n} diverges, the series diverges.
A geometric series has the form Σ ar^(n−1) = a + ar + ar² + ar³ + ...
Converges when |r| < 1, with sum s = a/(1 – r).
Diverges for all other values of r (including r = 1 and r = −1).
To use the formula, identify a (the first term) and r (the common ratio between consecutive terms).
Example: 2 + 2/3 + 2/9 + 2/27 + ... Here a = 2, r = 1/3. Sum = 2/(1 – 1/3) = 2/(2/3) = 3.
A telescoping series is one whose partial sums collapse after cancellation. Write out S_n, cancel adjacent terms, and take the limit of what remains.
Example: Σ [1/n – 1/(n+1)] = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + ... The partial sum is S_n = 1 – 1/(n+1), so the series converges to 1.
The series Σ 1/n = 1 + 1/2 + 1/3 + 1/4 + ... diverges, even though the terms approach zero. This is proved using the Integral Test (Section 10.3).
If lim(n→∞) a_n ≠ 0 (or the limit does not exist), then Σ a_n diverges.
Critical warning: this test can only prove divergence, never convergence. If lim a_n = 0, the test is inconclusive and you must use another test. The harmonic series is the classic example: lim 1/n = 0 but the series still diverges.
Suppose f(x) is continuous, positive, and decreasing on [k, ∞), and a_n = f(n). Then:
The series Σ a_n converges if and only if the improper integral ∫ from k to ∞ of f(x) dx converges.
The function does not need to be decreasing everywhere, only "eventually" (for all x > N for some N).
The integral and the series do not have the same value. The test only tells you whether they share the same convergence behaviour.
For f(x) = 1/x on [1, ∞): approximate the area under the curve with unit-width rectangles using left endpoints. The rectangle areas are 1 + 1/2 + 1/3 + ... = Σ 1/n, which overestimates the integral. Since ∫ 1/x dx = ln x diverges, the harmonic series must also diverge.
For f(x) = 1/x² on [1, ∞): use right-endpoint rectangles. The rectangle areas 1/4 + 1/9 + 1/16 + ... underestimate the integral. Since ∫ 1/x² dx = 1 converges, the series Σ 1/n² also converges (and its sum is less than 2).
The p-series Σ 1/n^p has a clean convergence rule:
Converges when p > 1.
Diverges when p ≤ 1.
The dividing line is p = 1, which gives the harmonic series.
Examples:
Σ 1/n^(5/2): converges (p = 2.5 > 1).
Σ 1/√n = Σ 1/n^(1/2): diverges (p = 0.5 ≤ 1).
Σ 1/(2n³) = (1/2)Σ 1/n³: converges (p = 3 > 1).
Σ n^(–e): converges (p = e ≈ 2.718 > 1).
Σ 1/(n² + 4): the integral ∫ 1/(x²+4) dx = (1/2) arctan(x/2), which converges. So the series converges.
Σ 1/(n ln n) for n ≥ 2: the integral ∫ 1/(x ln x) dx = ln(ln x), which diverges. So the series diverges.
Σ (ln n)/n for n ≥ 2: the integral ∫ (ln x)/x dx = (ln x)²/2, which diverges. So the series diverges.
Given a series Σ a_n with a_n ≥ 0:
If a_n ≤ b_n for all n (or eventually), and Σ b_n converges, then Σ a_n converges.
If a_n ≥ b_n for all n (or eventually), and Σ b_n diverges, then Σ a_n diverges.
The comparison series Σ b_n is usually a p-series or a geometric series whose behaviour you already know.
Strategy: find the dominating terms in the numerator and denominator of a_n; form b_n from their ratio.
Example 1: Σ 1/[(n+1) · 4^n]. For large n, the term (n+1) in the denominator makes a_n smaller than 1/4^n. Since Σ 1/4^n is a convergent geometric series (r = 1/4), and a_n < b_n = 1/4^n, the series converges by DCT.
Example 2: Σ n/(n – 8) for n ≥ 9. Choose b_n = 1/√n. Since n – 8 < n, we get n/(n−8) > n/n = 1/√n. The series Σ 1/√n diverges (p-series, p = 1/2), and a_n > b_n, so the series diverges by DCT.
Example 3: Σ (ln n)/n. Choose b_n = 1/n (the harmonic series, divergent). For n ≥ 3, ln n ≥ 1, so (ln n)/n ≥ 1/n. The inequality fails for n = 1 and n = 2, but that does not matter: convergence depends only on the tail.
Sometimes the inequality goes the wrong way. For Σ √n/(n³+1), the natural comparison b_n = 1/n^(5/2) gives √n/(n³+1) > 1/n^(5/2) (false, since n³+1 > n³). In these cases, use the Limit Comparison Test.
Given Σ a_n (a_n > 0) and a known series Σ b_n (b_n > 0):
If lim(n→∞) a_n/b_n = c, where 0 < c < ∞, then both series behave alike (both converge or both diverge).
If the limit is 0 and Σ b_n converges, then Σ a_n converges.
If the limit is ∞ and Σ b_n diverges, then Σ a_n diverges.
The LCT is often easier than the DCT because you only need to compute a limit, not establish an inequality.
Σ √n/(n³+1). Choose b_n = 1/n^(5/2). Then a_n/b_n = [√n/(n³+1)] / [1/n^(5/2)] = n³/(n³+1) → 1 as n → ∞. Since 1 is finite and positive, and Σ 1/n^(5/2) converges (p = 5/2 > 1), the original series also converges.
Geometric series sum
Σ ar^(n−1) = a/(1 – r), valid when |r| < 1
p-series convergence rule
Σ 1/n^p converges if and only if p > 1
Partial sum of a geometric series
S_n = a(1 – r^n)/(1 – r)
Geometric sequence limit
lim r^n = 0 if |r| < 1; = 1 if r = 1; diverges otherwise
The special limit
lim(n→∞) (1 + m/n)^n = e^m
Test for Divergence
If lim(n→∞) a_n ≠ 0, then Σ a_n diverges. (Converse is false.)
Students often think that if lim a_n = 0, the series must converge. It does not. The harmonic series is the classic counterexample: lim 1/n = 0 but Σ 1/n = ∞.
The Integral Test tells you whether the series converges, not what it converges to. The sum of the series and the value of the integral are different numbers.
When using the Direct Comparison Test, students sometimes set up the inequality backwards. Remember: to prove convergence, your series must be smaller than a convergent one. To prove divergence, it must be larger than a divergent one.
Students confuse sequences and series. A sequence is just a list; a series is a sum. The sequence {1/n} converges (to 0), but the series Σ 1/n diverges.
⚠️ Geometric series problems are nearly guaranteed on exams. You must be able to identify a, r, and apply the formula quickly.
⚠️ Know the p-series convergence rule cold: p > 1 converges, p ≤ 1 diverges. Many comparison test problems use a p-series as the benchmark.
⚠️ The Test for Divergence is often the first step. If lim a_n ≠ 0, you are done (diverges). If the limit is 0, you must try another test.
⚠️ Comparison test problems ask you to choose the right comparison series. Look at the dominant terms in the numerator and denominator to guide your choice.
⚠️ The limit (1 + 1/n)^n = e is a recurring exam problem and is also the foundation for continuously compounded interest.
True or False: The series Σ 1/n³ converges. True (p-series with p = 3 > 1).
Fill in the blank: A geometric series converges when ____ < 1. |r|.
True or False: If lim a_n = 0, then Σ a_n converges. False.
Fill in the blank: The harmonic series is the p-series with p = ____. 1.
True or False: The Direct Comparison Test requires an inequality that holds for all n starting from some point. True.
Q: Determine whether Σ (3^n)/(4^n) converges or diverges. If it converges, find the sum.
A: This is a geometric series with a = 3/4 and r = 3/4. Since |r| = 3/4 < 1, it converges. Sum = (3/4)/(1 – 3/4) = (3/4)/(1/4) = 3.
Q: Does the sequence a_n = (−1)^n · n/(n²+1) converge or diverge?
A: |a_n| = n/(n²+1) → 0. By the absolute value rule, since lim |a_n| = 0, the sequence converges to 0.
Q: Use the Limit Comparison Test to determine whether Σ (2n+1)/(n³+4) converges or diverges.
A: Dominant terms: 2n in the numerator, n³ in the denominator. Choose b_n = 1/n². Then a_n/b_n = (2n+1)n²/(n³+4) → 2 as n → ∞. Since 2 is finite and positive, and Σ 1/n² converges (p = 2 > 1), the original series converges.
Q: Does Σ 1/(n ln n) for n ≥ 2 converge or diverge?
A: Use the Integral Test. ∫ from 2 to ∞ of 1/(x ln x) dx. Let u = ln x, du = 1/x dx. The integral becomes ∫ 1/u du = ln|u| = ln(ln x), which diverges. So the series diverges.
Q: True or false: Σ 1/n^(1/3) converges.
A: False. This is a p-series with p = 1/3 < 1, so it diverges.
Sequences and their limits build directly on the limit concepts from Sections 2.2 and 2.6. L'Hôpital's Rule (Section 4.5) is essential for evaluating indeterminate-form sequence limits. The Integral Test links series convergence back to improper integrals from Chapter 8. Everything in this set of notes is prerequisite material for the Ratio Test, Root Test, Alternating Series Test, and Power Series (covered in Part 2 of these study notes).
Sequences, series, infinite series, partial sums, convergence, divergence, geometric series, common ratio, harmonic series, p-series, Integral Test, Direct Comparison Test, DCT, Limit Comparison Test, LCT, Test for Divergence, nth term test, Squeeze Theorem, bounded monotone sequence, telescoping series, Math 104, Penn State, Calculus 1, Chapter 10