Difficulty: Intermediate | Prerequisites: Infinite series fundamentals (partial sums, geometric series, Divergence Test)
This set of notes covers the main tools for deciding whether a series converges or diverges when you cannot find a closed-form sum. You should already be comfortable with the Divergence Test, geometric series, and telescoping series before working through this material. These convergence tests form the bulk of exam questions on series.
When a series is not geometric or telescoping, you need convergence tests. The Comparison Test and the Limit Comparison Test let you benchmark an unfamiliar series against one you already know. p-series (sum of 1/n^p) are the most common benchmarks: they converge for p > 1 and diverge for p <= 1. For series involving factorials or nth powers, ratio-style arguments are your go-to.
p-series
A series of the form sum from n = 1 to infinity of 1/n^p. Converges when p > 1, diverges when p <= 1. Think of it as the family of series that includes the harmonic series (p = 1, diverges) and the sum of 1/n^2 (p = 2, converges). This is the benchmark series you will compare against most often.
Comparison Test (Direct Comparison)
If 0 <= a_n <= b_n for all n, and the sum of b_n converges, then the sum of a_n converges. Conversely, if the sum of a_n diverges, then the sum of b_n diverges. In simple terms, a series smaller than a convergent series must also converge, and a series bigger than a divergent series must also diverge.
Limit Comparison Test
If lim (n to infinity) a_n / b_n = L, where 0 < L < infinity, then sum a_n and sum b_n either both converge or both diverge. Think of it as: if two series have terms that grow at the same rate, they share the same convergence behaviour.
Divergence Test (nth-term test)
If lim a_n is not 0, the series diverges. If lim a_n = 0, the test is inconclusive. Always check this first.
Ratio Test
If lim |a_(n+1) / a_n| = L, then: L < 1 means convergence, L > 1 means divergence, L = 1 is inconclusive. Particularly useful for series involving factorials or exponential terms.
The Comparison Test is your workhorse for series whose terms are positive and look "close to" a simpler series.
Example 1: sum of 1 / (n^(4/3) * n)
The term simplifies to 1 / n^(4/3 + 1) = 1 / n^(7/3). Since 7/3 > 1, this is a convergent p-series. Alternatively, note that 1 / (n^(4/3) * n) < 1 / n^2 for large n, and sum of 1/n^2 converges (p = 2 > 1). By the Comparison Test, the original series converges.
Example 2: sum of sin^2(n) / (n * sqrt(n))
Since sin^2(n) <= 1 for all n, we have:
sin^2(n) / (n * sqrt(n)) <= 1 / (n * sqrt(n)) = 1 / n^(3/2)
The series sum of 1/n^(3/2) is a p-series with p = 3/2 > 1, so it converges. By the Comparison Test, the original series converges.
The strategy: find a simpler series that is term-by-term larger (for proving convergence) or term-by-term smaller (for proving divergence), and whose behaviour you already know. p-series are almost always the right comparison target.
Key p-series reference:
p = 1 (harmonic series): diverges
p = 3/2: converges
p = 2: converges
p = 1/2: diverges
Any p > 1: converges
Any p <= 1: diverges
Example 3: sum from n = 2 of sin(1/n)
For small x, sin(x) is approximately x. So sin(1/n) behaves like 1/n for large n. Since the harmonic series (sum of 1/n) diverges, the Limit Comparison Test confirms that sum of sin(1/n) also diverges.
More precisely: lim [sin(1/n) / (1/n)] = 1 as n tends to infinity (a standard limit). Since 0 < 1 < infinity, the two series share convergence behaviour, and sum of 1/n diverges.
Example 4: sum from n = 2 of 1 / (n^2 - 1)
lim (n to infinity) of 1/(n^2 - 1) divided by 1/n^2 = lim n^2 / (n^2 - 1) = 1. By the Limit Comparison Test, since sum of 1/n^2 converges (p = 2 > 1), the original series also converges.
Alternatively, note that 1/(n^2 - 1) = 1/[(n-1)(n+1)], which can be decomposed by partial fractions into a telescoping series.
Example 5: sum from n = 1 of n^n / n!
Consider the ratio a_(n+1)/a_n or simply inspect the terms directly. We have n^n / n! = (n/1)(n/2)(n/3)...(n/n). Each factor n/k >= 1, and for k < n, n/k > 1. So n^n / n! >= n for all n >= 2.
Since lim (n to infinity) of n^n / n! = infinity (not 0), the Divergence Test tells us the series diverges.
Choosing the right test, a quick decision tree:
Terms do not go to zero? Divergence Test. Done.
Looks geometric (constant ratio)? Geometric series test.
Terms cancel in pairs after partial fractions? Telescoping.
Positive terms that look like 1/n^p? Direct or Limit Comparison with p-series.
Factorials or nth powers? Try the Ratio Test.
None of the above? Try the Integral Test or Root Test (covered in later notes).
p-series: sum of 1/n^p converges if and only if p > 1.
Comparison Test: 0 <= a_n <= b_n and sum b_n converges implies sum a_n converges.
Limit Comparison Test: if lim (a_n / b_n) = L with 0 < L < infinity, then sum a_n and sum b_n share convergence behaviour.
Divergence Test: lim a_n not equal to 0 implies sum a_n diverges.
Ratio Test: lim |a_(n+1) / a_n| = L. L < 1: converges. L > 1: diverges. L = 1: inconclusive.
Useful approximation: for small x, sin(x) is approximately x. This means sin(1/n) behaves like 1/n for large n, which is critical for limit comparison arguments.
Students often try to use the Comparison Test with the wrong inequality direction. To prove convergence, your series must be smaller than a known convergent series. To prove divergence, it must be larger than a known divergent series. Getting this backwards proves nothing.
Students confuse the Comparison Test with the Limit Comparison Test. The direct comparison requires an inequality (a_n <= b_n). The limit comparison only requires that the ratio a_n/b_n approaches a finite positive number. When you cannot easily establish an inequality, the Limit Comparison Test is usually easier.
A common error with sin(1/n): students see "sin" and assume the series oscillates or that sin^2 terms make it diverge. For large n, sin(1/n) is approximately 1/n, so it behaves like the harmonic series and diverges. The sin function does not automatically mean oscillation when its argument is shrinking.
Students sometimes claim n^n / n! converges because "factorials grow fast." Factorials do grow fast, but n^n grows faster. Each factor in the product n * n * n * ... * n (n times) divided by 1 * 2 * 3 * ... * n is at least 1, and most are strictly greater. The terms blow up.
⚠️ The Comparison Test with p-series is one of the most frequently tested skills. You will almost certainly see a problem asking you to determine convergence of a series like 1 / (n^2 + 5n) or sin^2(n) / n^(3/2). Know how to bound terms and identify the dominant power of n.
⚠️ The Limit Comparison Test is the cleaner alternative when the inequality is hard to establish. Exams often present series where direct comparison is awkward but the limit comparison is straightforward.
⚠️ The sin(1/n) divergence via limit comparison with 1/n is a classic exam question. Memorise the fact that lim [sin(x)/x] = 1 as x tends to 0, and know how to apply it.
⚠️ Be ready to explain why n^n / n! diverges. Exam questions testing the Divergence Test often use terms that look intimidating but whose limit is clearly nonzero (or infinite).
True or false: the series sum of 1/n^(1/2) converges.
Fill in the blank: the Comparison Test can prove convergence only when your series is ______ than a known convergent series.
True or false: sum of sin(1/n) from n = 1 to infinity converges.
Fill in the blank: in the Limit Comparison Test, the ratio a_n/b_n must approach a value L that is ______ and ______.
True or false: sum of n^n / n! converges.
(Answers: 1. False, p = 1/2 <= 1. 2. Smaller (term by term). 3. False, it diverges by limit comparison with the harmonic series. 4. Positive and finite (0 < L < infinity). 5. False, the terms tend to infinity, so it diverges by the Divergence Test.)
Q: Determine whether the series sum from n = 1 to infinity of 1 / (n^2 + 4) converges or diverges.
A: For large n, 1/(n^2 + 4) < 1/n^2, and sum of 1/n^2 converges (p = 2 > 1). By the Comparison Test, the series converges.
Q: Does the series sum from n = 1 to infinity of n / (n^2 + 1) converge?
A: For large n, n/(n^2 + 1) behaves like n/n^2 = 1/n. lim [n/(n^2+1)] / [1/n] = lim n^2/(n^2+1) = 1. By the Limit Comparison Test with the harmonic series, this series diverges.
Q: Use the Comparison Test to show that sum of sin^2(n) / n^2 converges.
A: sin^2(n) <= 1, so sin^2(n)/n^2 <= 1/n^2. Sum of 1/n^2 converges (p-series, p = 2). By the Comparison Test, the original series converges.
Q: Determine whether sum from n = 1 to infinity of (2^n) / (n!) converges or diverges.
A: Apply the Ratio Test: a_(n+1)/a_n = 2^(n+1)/(n+1)! * n!/2^n = 2/(n+1). As n tends to infinity, this ratio tends to 0, which is less than 1. The series converges.
Q: Why does sum from n = 2 to infinity of sin(1/n) diverge even though sin(1/n) tends to 0?
A: The Divergence Test only catches series whose terms do not go to 0. Here the terms do go to 0, so the test is silent. The divergence comes from the Limit Comparison Test: sin(1/n) / (1/n) tends to 1, and the harmonic series diverges, so sin(1/n) diverges too. Terms going to 0 is necessary but not sufficient for convergence.
The comparison tests build directly on your knowledge of p-series and geometric series from the fundamentals notes. If you are not fluent with those benchmark series, the comparison tests will feel like guessing.
The Ratio Test connects forward to power series and Taylor series, where it determines the radius of convergence. If you will be covering Taylor series later in the course, the Ratio Test is the single most important convergence tool to master now.
The small-angle approximation sin(x) approximately equals x is a preview of Taylor series ideas. You will see it again when studying linearisation and first-order Taylor polynomials.
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