Difficulty: Intermediate to Advanced | Prerequisites: Limits at infinity, improper integrals, comfort with algebraic manipulation of rational expressions.
Sequences and series form the bridge between finite sums and infinite processes. This is the part of integral calculus where you ask: does an infinite sum have a finite value, and if so, what is it? The convergence tests are your decision tree. Each test handles a different structural pattern in the series, and exam success depends on recognising which test to apply. This material is the foundation for power series, Taylor series, and Fourier series in later courses.
A sequence converges if its terms approach a finite limit. A series converges if its partial sums approach a finite limit. You have about half a dozen convergence tests, each suited to a particular type of series. Picking the right test is the skill the exam is testing.
Sequence
An ordered list of numbers a_1, a_2, a_3, ... defined by a formula. A sequence converges if lim a_n as n approaches infinity exists and is finite.
Series
The sum of the terms of a sequence: sum from k = 1 to infinity of a_k. A series converges if the sequence of partial sums converges.
Geometric Series
A series of the form sum of a r^k. Converges if and only if |r| < 1, in which case the sum is a/(1 - r). Think of it as the workhorse for repeating decimals and many closed-form sums.
p-Series
The series sum of 1/k^p. Converges if p > 1, diverges if p is less than or equal to 1. The harmonic series (p = 1) is the classic divergent example.
Divergence Test (nth-Term Test)
If lim a_n is not equal to 0, the series diverges. This test can only prove divergence, never convergence.
Comparison Test
If 0 is less than or equal to a_k is less than or equal to b_k for all k and sum of b_k converges, then sum of a_k converges. If a_k is greater than or equal to c_k is greater than or equal to 0 and sum of c_k diverges, then sum of a_k diverges.
Integral Test
If f(x) is continuous, positive and decreasing on [1, infinity) and a_k = f(k), then sum of a_k converges if and only if the improper integral of f(x) from 1 to infinity converges.
Alternating Series Test (Leibniz Test)
A series of the form sum of (-1)^n b_n converges if b_n is eventually decreasing and lim b_n = 0.
Ratio Test
Compute L = lim |a_(n+1)/a_n|. If L < 1, the series converges absolutely. If L > 1, it diverges. If L = 1, the test is inconclusive. Particularly useful for series involving factorials or exponentials.
Root Test
Compute L = lim (|a_k|)^(1/k). Same conclusions as the ratio test. Best for series where the general term is raised to the kth power.
Absolutely Convergent
A series is absolutely convergent if the series of absolute values converges. Absolute convergence implies convergence.
Conditionally Convergent
A series that converges but does not converge absolutely. The alternating harmonic series is the standard example.
To find the limit of a rational-type sequence, divide numerator and denominator by the highest power of n.
Example: a_n = 14n^2/(7n+3) - (2n^2+8)/(n+1). Combine over a common denominator to get (8n^2 - 56n - 24)/(7n^2 + 10n + 3). Divide through by n^2: limit is 8/7.
A repeating decimal can be written as a whole-number part plus a geometric series.
Example: 2.959595... = 2 + 95/100 + 95/10000 + ... = 2 + (95/100) times 1/(1 - 1/100) = 2 + 95/99 = 293/99.
The common ratio is 1/100 (or 1/10^2 for a two-digit repeat).
If lim a_n is not 0, the series diverges. Full stop.
Example: sum of (-1)^n (6n-4)/(7n-5). The limit of (6n-4)/(7n-5) is 6/7, which is not 0. The series diverges.
This test never proves convergence. A limit of 0 only means this test is inconclusive.
Compare your series term-by-term with a known series.
Example: sum of (k+9)/k^2. Since (k+9)/k^2 > k/k^2 = 1/k > 0, and the harmonic series sum of 1/k diverges, the comparison test gives divergence.
Works when the terms come from a continuous, positive, decreasing function.
Example: sum of 1/(k(ln(4k))^2). Set f(x) = 1/(x(ln(4x))^2). Substitute u = ln(4x). The improper integral converges (it evaluates to 1/ln(4)), so the series converges.
For sum of (-1)^n b_n: check that b_n is eventually decreasing and that lim b_n = 0.
Example: sum of (-1)^(n+6)/n^(1/4). Since 1/n^(1/4) is decreasing and approaches 0, the series converges by the alternating series test.
However, sum of 1/n^(1/4) diverges (p-series with p = 1/4 < 1), so the convergence is only conditional.
Best for factorials and exponential terms.
Example: sum of (2n+1)/(2n)!. Compute |a_(n+1)/a_n| = (2n+3)/((2n+1)(2n+2)) times (2n+3)/(2n+1). The limit is 0, so the series is absolutely convergent.
Example: sum of e^(-3k) times k!. The ratio (k+1)/e^3 approaches infinity, so the series diverges.
Best when the general term is something raised to the kth power.
Example: sum of (6^(1/k) - 1)^k. The kth root is 6^(1/k) - 1, which approaches 1 - 1 = 0 as k approaches infinity. Since 0 < 1, the series converges.
Absolute convergence: sum of |a_n| converges. This automatically means sum of a_n converges too.
Conditional convergence: sum of a_n converges, but sum of |a_n| diverges.
Example: sum of (-1)^(n+6)/n^(1/4) converges conditionally. It passes the alternating series test, but the absolute-value series (a p-series with p = 1/4) diverges.
\sum_{k=0}^{\infty} ar^k = \frac{a}{1-r}, \quad |r| < 1\sum_{k=1}^{\infty} \frac{1}{k^p} \text{ converges iff } p > 1L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \quad \text{(Ratio Test)}L = \lim_{k \to \infty} |a_k|^{1/k} \quad \text{(Root Test)}Geometric series model compound interest, drug dosage accumulation and signal attenuation in communications. The ratio and root tests are essential when studying the radius of convergence of power series, which arise in solving differential equations and in numerical approximation of functions. Convergence tests appear in any field that uses infinite sums: Fourier analysis for signal processing, perturbation theory in physics, and numerical methods in engineering.
The divergence test only works one way. If the limit of a_n is 0, you cannot conclude convergence. The harmonic series has terms approaching 0 but diverges.
Students frequently apply the wrong test. The ratio test is ideal for factorials and exponentials; the root test for terms raised to the kth power; the comparison test for rational functions of k. Applying the ratio test to a p-series, for instance, always gives L = 1 (inconclusive).
With the alternating series test, both conditions matter: b_n must be eventually decreasing and must approach 0. Checking only one condition is insufficient.
Students confuse absolute and conditional convergence. A conditionally convergent series converges, but its absolute-value version diverges. Rearranging the terms of a conditionally convergent series can change its sum (Riemann rearrangement theorem).
Expect multiple convergence-test questions. The exam tests your ability to pick the right test as much as your ability to execute it.
Repeating-decimal-to-fraction via geometric series is a common exam question. Know the pattern: digit block divided by (10^d - 1) where d is the number of repeating digits.
Questions about absolute vs conditional convergence often combine the alternating series test with the p-series test. Be ready to check both.
The divergence test is the fastest test. Always check it first: if the terms do not approach 0, you are done.
True or False: If lim a_n = 0, then sum of a_n converges. _____ (False: the harmonic series is a counterexample.)
Fill in the blank: A geometric series with common ratio r converges if and only if |r| ____ 1. (is less than)
True or False: The series sum of (-1)^n / sqrt(n) is absolutely convergent. _____ (False: it is conditionally convergent.)
Fill in the blank: The repeating decimal 0.787878... equals ____ / 99. (78)
True or False: The ratio test is inconclusive (L = 1) for p-series. _____ (True.)
Q: Find the limit of the sequence a_n = 14n^2/(7n+3) - (2n^2+8)/(n+1).
A: Combine over a common denominator to get (8n^2 - 56n - 24)/(7n^2 + 10n + 3). Divide by n^2: limit = 8/7.
Q: Express the repeating decimal 2.959595... as a fraction.
A: Write as 2 + 95/100 + 95/10000 + ... = 2 + (95/100)(1/(1 - 1/100)) = 2 + 95/99 = 293/99.
Q: Does sum of 1/(k(ln(4k))^2) from k = 1 to infinity converge or diverge?
A: Use the integral test. The improper integral of 1/(x(ln(4x))^2) from 1 to infinity converges (substitute u = ln(4x)), so the series converges.
Q: Does sum of (k+9)/k^2 converge or diverge?
A: Since (k+9)/k^2 > 1/k for all k at least 1, and the harmonic series diverges, the comparison test gives divergence.
Q: Is sum of (-1)^(n+6)/n^(1/4) absolutely convergent, conditionally convergent, or divergent?
A: Conditionally convergent. The alternating series test confirms convergence (1/n^(1/4) is decreasing, limit is 0). But the absolute-value series sum of 1/n^(1/4) is a p-series with p = 1/4 < 1, which diverges.
Q: Determine convergence of sum of (6^(1/k) - 1)^k.
A: By the root test, |a_k|^(1/k) = 6^(1/k) - 1, which approaches 0 as k approaches infinity. Since 0 < 1, the series converges.
Convergence tests are the prerequisite for power series and Taylor series, where you use the ratio test to find the radius of convergence. The integral test ties this material back to improper integrals. Geometric series appear again in differential equations (Laplace transforms) and in probability (geometric distributions).
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