Difficulty: Intermediate | Prerequisites: Calculus 1 (limits, derivatives, integrals), familiarity with basic sequences
Infinite series extend the idea of adding finitely many numbers to adding infinitely many. The central question is whether that infinite sum settles down to a finite value (converges) or grows without bound (diverges). This matters because many functions in mathematics, physics, and engineering are represented as infinite series (Taylor series, Fourier series), and knowing whether those representations are valid depends on convergence.
This section covers the ten standard tests for determining convergence or divergence. Choosing the right test is a skill in itself, and exams routinely ask you to justify which test you used and why.
An infinite series either converges to a finite number or diverges. There are ten standard tests to determine which. Most exam problems boil down to recognising which test applies, checking its conditions, and drawing the correct conclusion, with special care around the cases where a test is inconclusive.
Infinite series
The sum a_1 + a_2 + a_3 + ..., written as the sum from n=1 to infinity of a_n. Think of it as an attempt to add up infinitely many numbers.
Partial sum (s_n)
The sum of the first n terms: s_n = a_1 + a_2 + ... + a_n. In simple terms, this is where you stop adding and check the running total.
Convergence
A series converges if the sequence of partial sums approaches a finite limit s as n goes to infinity. In simple terms, the running total settles down to a specific number.
Divergence
A series diverges if the partial sums do not approach a finite limit. The total either grows without bound, oscillates, or otherwise fails to settle.
Geometric series
A series of the form a + ar + ar^2 + ar^3 + ..., where r is the common ratio. It converges to a/(1-r) when |r| < 1 and diverges when |r| >= 1.
P-series
A series of the form sum of 1/n^p. It converges when p > 1 and diverges when p <= 1.
Harmonic series
The p-series with p = 1: sum of 1/n = 1 + 1/2 + 1/3 + 1/4 + ... This series diverges, even though its terms go to zero. A classic trap.
An infinite series is written as sum from n=1 to infinity of a_n = a_1 + a_2 + a_3 + ...
The nth partial sum is s_n = sum from i=1 to n of a_i = a_1 + a_2 + ... + a_n.
If lim (n to infinity) s_n = s, where s is a real number, then the infinite sum equals s and the series converges. If the limit does not exist or is not finite, the series diverges.
The distinction between the terms a_n and the partial sums s_n is critical. A series can have terms that go to zero and still diverge (the harmonic series is the standard example).
Form: sum from n=0 to infinity of ar^n = a + ar + ar^2 + ar^3 + ...
Convergence condition
Converges to a / (1 - r) when |r| < 1
Diverges when |r| >= 1
The key is the common ratio r. If each term is less than the previous one in absolute value (|r| < 1), the total settles to a finite number. If not, it blows up.
Example: sum of (1/2)^n from n=0 to infinity = 1 / (1 - 1/2) = 2.
Example: sum of 2^n from n=0 to infinity diverges because r = 2 and |r| >= 1.
Form: sum from n=1 to infinity of 1/n^p
Convergence condition
Converges when p > 1
Diverges when p <= 1
The boundary case p = 1 gives the harmonic series: 1 + 1/2 + 1/3 + 1/4 + ..., which diverges. This is worth memorising outright, because exams test whether you know it diverges despite its terms going to zero.
Examples:
sum of 1/n^2 converges (p = 2 > 1)
sum of 1/n^(1/2) = sum of 1/sqrt(n) diverges (p = 1/2 <= 1)
sum of 1/n diverges (p = 1, the harmonic series)
Divergence test (test 4 in the cheat sheet)
If lim (n to infinity) a_n is not equal to 0, the series diverges. If lim a_n = 0, the test is inconclusive and you must use another test. This is the first test to try. It can only prove divergence, never convergence.
Integral test (test 5)
Given sum of a_n where a_n > 0 and a_n is decreasing:
If the integral from c to infinity of a_n dn converges, the series converges.
If the integral diverges, the series diverges.
The integral and the series either both converge or both diverge. Note: the integral gives you convergence or divergence, but not the exact sum of the series.
Ratio test (test 6)
Given sum of a_n with a_n > 0, compute p = lim (n to infinity) |a_(n+1) / a_n|.
If p < 1, the series converges.
If p > 1, the series diverges.
If p = 1, the test is inconclusive.
The ratio test works especially well for series involving factorials (n!) or exponentials (r^n), where successive terms have a clean ratio.
Root test (test 7)
Given sum of a_n with a_n > 0, compute p = lim (n to infinity) (a_n)^(1/n).
If p < 1, the series converges.
If p > 1, the series diverges.
If p = 1, the test is inconclusive.
The root test is most useful when a_n is already raised to the nth power, such as (expression)^n. In many cases the ratio test and root test give the same answer; pick whichever simplifies more cleanly.
Limit comparison test (test 8)
Given two series with a_n > 0 and b_n > 0, compute p = lim (n to infinity) a_n / b_n.
If p is a finite positive number (p > 0, p finite), then both series either converge or both diverge. Choose b_n to be a series whose behaviour you already know (typically a p-series or geometric series).
Direct comparison test (test 9)
Given a_n >= 0, b_n >= 0, and a_n <= b_n for all n:
If b_n converges, then a_n converges.
If a_n diverges, then b_n diverges.
In other words, a convergent series bounds everything below it, and a divergent series drags everything above it. You need the inequality to run the right way.
Alternating series test (test 10)
Given a series of the form sum of (-1)^n a_n, with a_n > 0:
If a_(n+1) < a_n (the terms are decreasing) and lim (n to infinity) a_n = 0, the series converges.
This test only works for series that alternate in sign. It tells you the series converges, but does not give you the sum. The classic example is the alternating harmonic series: sum of (-1)^(n+1) / n = 1 - 1/2 + 1/3 - 1/4 + ..., which converges (to ln 2), even though the ordinary harmonic series diverges.
Students often assume that if a_n goes to zero, the series must converge. This is false. The harmonic series (1/n) has terms going to zero and still diverges. The divergence test only works in one direction: if the terms do not go to zero, the series definitely diverges.
Students often confuse the ratio test and the root test outcomes when p = 1. Both tests are inconclusive at p = 1. You cannot conclude convergence or divergence, and you must try a different test.
Students often apply the direct comparison test with the inequality running the wrong way. If your known series diverges, you need a_n >= b_n (your series is bigger) to conclude divergence. If your known series converges, you need a_n <= b_n (your series is smaller) to conclude convergence.
Students often forget to check the conditions of the alternating series test. Both conditions are required: the terms must be decreasing and must approach zero. Missing either one invalidates the test.
Expect at least one problem asking you to determine convergence or divergence and to state which test you used. Justification matters.
The geometric series formula is commonly tested as a standalone calculation: find the sum of a given geometric series. Know the formula cold.
The harmonic series is a perennial exam trap. Be ready for it to appear as a comparison target or as a trick question.
The ratio test is the go-to for series with factorials or exponentials. If you see n! or r^n in the terms, start with the ratio test.
Alternating series problems often ask whether a series converges absolutely, conditionally, or diverges. Know the difference: absolute convergence means the series of absolute values converges; conditional convergence means the series converges but the series of absolute values diverges.
True or false: If lim (n to infinity) a_n = 0, the series sum of a_n converges.
False. The terms going to zero is necessary but not sufficient. The harmonic series is the classic counterexample.
Fill in the blank: A geometric series with |r| < 1 converges to ____.
a / (1 - r).
True or false: The ratio test is conclusive when p = 1.
False. When p = 1, the ratio test gives no conclusion.
Fill in the blank: The p-series sum of 1/n^p converges when p ____.
is greater than 1 (p > 1).
True or false: The alternating series test can prove divergence.
False. It can only prove convergence. If its conditions are not met, you need a different test.
Q: Does the series sum from n=1 to infinity of n/(n+1) converge or diverge?
A: Diverges. lim (n to infinity) n/(n+1) = 1, which is not zero, so the series diverges by the divergence test.
Q: Find the sum of the geometric series 5 + 5/3 + 5/9 + 5/27 + ...
A: Here a = 5 and r = 1/3. Since |r| < 1, the sum is a/(1-r) = 5 / (1 - 1/3) = 5 / (2/3) = 15/2.
Q: Does sum from n=1 to infinity of 1/n^3 converge or diverge?
A: Converges. This is a p-series with p = 3 > 1.
Q: Use the ratio test to determine whether sum from n=1 to infinity of n! / 3^n converges or diverges.
A: Compute lim (n to infinity) |a_(n+1)/a_n| = lim [(n+1)! / 3^(n+1)] / [n! / 3^n] = lim (n+1)/3 = infinity. Since p > 1, the series diverges.
Q: Does the alternating series sum from n=1 to infinity of (-1)^(n+1) / n converge?
A: Yes. The terms 1/n are positive, decreasing, and approach zero. By the alternating series test, the series converges. (It converges conditionally, since the harmonic series of absolute values diverges.)
Infinite series lead directly into Taylor and Maclaurin series, where you represent functions as power series and use the ratio test to find their radius of convergence. This is one of the most heavily tested topics in Calc 2.
The integral test connects series back to the integration techniques from Calc 1. If you can evaluate an improper integral, you can determine convergence of the corresponding series.
In differential equations and physics, series solutions are used when a differential equation cannot be solved in closed form. Fourier series, used throughout engineering, depend on the same convergence ideas.
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