Source: Integral Calculus, University of Texas at Austin
Tags: ratio test, root test, geometric series, summation notation, trigonometric series, Calculus II
The Ratio Test and Root Test are your go-to tools when a series involves factorials, exponentials, or terms raised to the nth power. Geometric series with trigonometric arguments round out this set, testing whether you can identify when a geometric series converges and compute its sum. This section also covers rewriting series in summation notation, a skill that exams test directly.
Difficulty: Intermediate to Advanced
Prerequisites: alternating series test, absolute and conditional convergence, geometric series formula, factorials, basic trigonometric identities (sin^2 + cos^2 = 1).
The Ratio Test and Root Test are the power tools of series convergence. Where comparison tests require you to find the right benchmark series, these tests work mechanically: compute a limit, and the answer falls out. They are especially effective for series involving factorials (Ratio Test) and nth-power expressions (Root Test).
Geometric series tie everything together. They are the simplest convergent series, and recognising a series as geometric (or comparing to one) is a skill you will use throughout the rest of calculus and into differential equations.
If you are behind, review geometric series (what is the common ratio, when does it converge, what is the sum) and make sure you are comfortable with factorial arithmetic (how n! and (2n)! relate to each other).
Ratio Test
Compute L = lim (n -> infinity) |a_(n+1) / a_n|. If L < 1, the series converges absolutely. If L > 1 (or L = infinity), the series diverges. If L = 1, the test is inconclusive. Think of it as checking whether each term shrinks by a consistent factor.
Root Test (Cauchy Root Test)
Compute rho = lim (n -> infinity) |a_n|^(1/n). If rho < 1, the series converges absolutely. If rho > 1, the series diverges. If rho = 1, the test is inconclusive. In simple terms, it extracts the "effective base" of the nth term.
Geometric series
A series of the form sum of ar^k. Converges when |r| < 1, with sum = a/(1-r). The common ratio r is the key: everything depends on its absolute value.
Summation notation (sigma notation)
A compact way to write a series using the sum symbol. The challenge is identifying the pattern, the starting index, and expressing the general term. Think of it as reverse-engineering a formula from a sequence of terms.
Common ratio
The fixed multiplier between consecutive terms in a geometric series. For sum of (sin^2(theta))^k, the common ratio is sin^2(theta).
The Ratio Test works best when a_n involves factorials or products. Compute the ratio |a_(n+1) / a_n|, simplify, and take the limit.
For the series sum of (n!)^2 * 3^n / (2n)!: the ratio of consecutive terms is
((n+1)!)^2 * 3^(n+1) / (2(n+1))!) / ((n!)^2 * 3^n / (2n)!) = 3(n+1)^2 / ((2n+2)(2n+1)).
As n approaches infinity, this approaches 3/4 < 1, so the series converges.
For the series sum of (6n)^n / n!: the ratio grows without bound (the numerator's growth outpaces the factorial). The series diverges.
For series (B) in problem 5, sum of (6n/(3n+5))^n * (2/3)^n: rewrite as sum of (6n * 2 / (3(3n+5)))^n = sum of (4n/(3n+5))^n. As n grows, 4n/(3n+5) approaches 4/3 > 1, so the terms do not approach zero. The series diverges.
The Root Test is ideal when a_n is already raised to the nth power (or 2n, etc.). Compute rho = lim |a_n|^(1/n).
For the series sum of ((4n+1)/(6n))^(2n): take the nth root to get ((4n+1)/(6n))^2. As n approaches infinity, (4n+1)/(6n) approaches 2/3, so rho = (2/3)^2 = 4/9 < 1. The series converges.
For series involving 3^n/(n+7)^n: the nth root gives 3/(n+7), which approaches 0. So rho = 0 < 1, and the series converges.
For sum of (4n/(7n+3))^n * (3/2)^n: this equals sum of (4n * 3 / (2(7n+3)))^n = sum of (6n/(7n+3))^n. The base approaches 6/7 < 1, so rho = 6/7 < 1 and the series converges.
For sum of n! * (4/n)^n: using Stirling's approximation or the Root Test, the nth root of n! * (4/n)^n grows without bound. The series diverges.
Series I: sum of n^n / 3^(2n+1) = sum of (n/9)^n / 3. The base n/9 grows without bound, so rho = infinity. Diverges.
Series II: sum of (1/5^n) * (2n)! / (n!)^2. The central binomial coefficient (2n)!/(n!)^2 grows like 4^n / sqrt(pi*n) (by Stirling). So the terms behave like 4^n / (5^n * sqrt(n)) = (4/5)^n / sqrt(n). Since 4/5 < 1, the series converges.
Answer: only series II converges.
The series sum from k=1 of (sin^2(theta))^k is geometric with first term sin^2(theta) and common ratio r = sin^2(theta).
It converges when |sin^2(theta)| < 1, which happens whenever sin(theta) is not equal to plus or minus 1, i.e. theta is not equal to pi/2 or 3pi/2 (within [0, 2pi)).
At theta = 5pi/3: sin(5pi/3) = -sqrt(3)/2, so sin^2(5pi/3) = 3/4. The sum is (3/4) / (1 - 3/4) = (3/4) / (1/4) = 3.
When the series converges, its sum is sin^2(theta) / (1 - sin^2(theta)) = sin^2(theta) / cos^2(theta) = tan^2(theta).
The series does not converge at theta = pi/2 and theta = 3pi/2 (where sin^2(theta) = 1, making the common ratio equal to 1).
The series 7(1/2)^2 * sin(3)/5 - 7(1/2)^3 * sin(4)/6 + 7(1/2)^4 * sin(5)/7 + ...
Identify the pattern:
Powers of (1/2): 2, 3, 4, ... so the exponent is k for k starting at 2, or (k-1) for k starting at 3.
Signs alternate starting negative at the second term: (-1)^(k-1) with appropriate indexing, or equivalently (-1/2)^(k-1).
sin arguments: 3, 4, 5, ... = k for k starting at 3.
Denominators: 5, 6, 7, ... = k+2 for k starting at 3.
Putting it together: sum from k=3 to infinity of (-1/2)^(k-1) * 7 sin(k) / (k+2).
The answer is: sum = sum from k=3 to infinity of (-1/2)^(k-1) * 7 sin(k) / (k+2).
Ratio Test
L = lim (n -> infinity) |a_(n+1) / a_n|. Converges absolutely if L < 1. Diverges if L > 1. Inconclusive if L = 1.
Root Test
rho = lim (n -> infinity) |a_n|^(1/n). Converges absolutely if rho < 1. Diverges if rho > 1. Inconclusive if rho = 1.
Geometric series sum
sum from k=0 of r^k = 1/(1-r) when |r| < 1. Starting from k=1: sum = r/(1-r).
Stirling's approximation (useful reference)
n! is approximately sqrt(2pin) * (n/e)^n. The central binomial coefficient (2n)!/(n!)^2 is approximately 4^n / sqrt(pi*n).
Trig identity
sin^2(theta) / cos^2(theta) = tan^2(theta). Also, 1 - sin^2(theta) = cos^2(theta).
Students often apply the Ratio Test to series where the Root Test is simpler (and vice versa). Rule of thumb: if the general term has an nth power, try the Root Test first. If it has factorials, try the Ratio Test first.
A common error with the Root Test is forgetting to take the nth root of the entire term. For ((4n+1)/(6n))^(2n), the nth root is ((4n+1)/(6n))^2, not (4n+1)/(6n).
Students often think that the geometric series sum from k=1 of r^k equals 1/(1-r). It does not. That formula is for the sum starting from k=0. Starting from k=1, the sum is r/(1-r).
When rewriting in summation notation, students frequently get the starting index wrong or use the wrong sign pattern. Always check: does your formula reproduce the first two or three terms of the given series?
⚠️ The Ratio and Root Tests are the standard approach for series with factorials and nth powers. Exams will give you series specifically designed for one or the other.
⚠️ For the Root Test, remember that rho = (2/3)^2 = 4/9, not 2/3. The exponent matters.
⚠️ Geometric series with trig arguments: the common ratio is sin^2(theta), and the series diverges exactly when sin^2(theta) = 1, i.e. at theta = pi/2 and 3pi/2. Exams test whether you can identify these values.
⚠️ Summation notation problems test pattern recognition. The safest approach is to write a table of the first few terms, identify what changes (index, sign, exponent, argument), and build the formula from there.
Fill in the blank: The Ratio Test says a series converges absolutely when L = lim |a_(n+1)/a_n| is less than ______.
True or False: The Root Test is inconclusive when rho = 1.
True.
Fill in the blank: The geometric series sum from k=1 of r^k converges to ______ when |r| < 1.
r/(1-r).
True or False: sin^2(theta) = 1 when theta = 0.
False. sin(0) = 0, so sin^2(0) = 0.
Fill in the blank: The series sum of (sin^2(theta))^k does not converge when theta = ______ or ______ (in [0, 2pi)).
pi/2 or 3pi/2.
Q: Does the series sum of (n!)^2 * 3^n / (2n)! converge or diverge?
A: Converges. The Ratio Test gives L = 3/4 < 1.
Q: Compute rho for the series sum of ((4n+1)/(6n))^(2n).
A: rho = (2/3)^2 = 4/9.
Q: What is the sum of the geometric series sum from k=1 of (sin^2(theta))^k at theta = 5pi/3?
A: sin^2(5pi/3) = 3/4. Sum = (3/4)/(1 - 3/4) = 3.
Q: Find the general sum of sum from k=1 of (sin^2(theta))^k whenever it converges.
A: sin^2(theta)/(1 - sin^2(theta)) = sin^2(theta)/cos^2(theta) = tan^2(theta).
Q: At which values of theta in [0, 2pi) does sum of (sin^2(theta))^k not converge?
A: theta = pi/2 and theta = 3pi/2, where sin^2(theta) = 1.
Q: Which of the series (A) sum of (6n)^n / n!, (B) sum of (6n/(3n+5))^n * (2/3)^n diverge?
A: Both diverge. (A) diverges because the terms grow without bound. (B) simplifies to terms with base approaching 4/3 > 1.
Q: Rewrite 7(1/2)^2 * sin(3)/5 - 7(1/2)^3 * sin(4)/6 + ... in summation notation.
A: sum from k=3 to infinity of (-1/2)^(k-1) * 7 sin(k) / (k+2).
The Ratio and Root Tests are the primary tools for finding the radius of convergence of power series, which is the next major topic in most Calculus II courses. The geometric series formula reappears when you derive Taylor series for 1/(1-x) and related functions.
Summation notation is a prerequisite for everything that follows: Taylor series, Fourier series, and any work with sequences and series in higher mathematics.
Ratio test, root test, Cauchy root test, geometric series, common ratio, summation notation, sigma notation, factorial, Stirling's approximation, central binomial coefficient, trig series, sin squared, tan squared, convergence radius, power series, Calculus II, integral calculus, UT Austin, HW 11