D’Alembert’s Ratio Test of Convergence of Series
The D’Alembert Ratio Test is one of the most useful tools for determining whether an infinite series converges or diverges. If you’re studying real analysis or calculus, you’ll use this constantly.
Statement of D’Alembert’s Ratio Test
A series \( \sum u_n \) of positive terms is convergent if, from and after some fixed term, \( \dfrac{u_{n+1}}{u_n} < r < 1 \), where \( r \) is a fixed number. The series is divergent if \( \dfrac{u_{n+1}}{u_n} > 1 \) from and after some fixed term.
D’Alembert’s Test is also known as the ratio test of convergence of a series.
Theorem
Let \( \displaystyle \sum_{n=1}^\infty a_n \) be a series of real numbers in \( \mathbb{R} \), or a series of complex numbers in \( \mathbb{C} \).
Let the sequence \( \{a_n\} \) satisfy:
$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = l$$
- If \( l > 1 \), then \( \displaystyle \sum_{n=1}^\infty a_n \) diverges.
- If \( l < 1 \), then \( \displaystyle \sum_{n=1}^\infty a_n \) converges absolutely.
- If \( l = 1 \), the test is inconclusive.
Key Definitions
(Definition 1: Convergence) An infinite series \( \sum u_n \), i.e., \( u_1 + u_2 + u_3 + \ldots + u_n \), is said to be convergent if \( S_n \), the sum of its first \( n \) terms, tends to a finite limit \( S \) as \( n \) tends to infinity.
We call \( S \) the sum of the series, and write:
$$S = \lim_{n \to \infty} S_n$$
More precisely, an infinite series \( \sum u_n \) converges to a sum \( S \) if for any given positive number \( \epsilon \), however small, there exists a positive integer \( n_0 \) such that \( |S_n – S| < \epsilon \) for all \( n \ge n_0 \).
(Definition 2: Divergence) If \( S_n \to \pm\infty \) as \( n \to \infty \), the series is said to be divergent. That is, \( \sum u_n \) is divergent if for every given positive number \( \lambda \), however large, there exists a positive integer \( n_0 \) such that \( |S_n| > \lambda \) for all \( n \ge n_0 \).
(Definition 3: Oscillation) If \( S_n \) does not tend to a finite limit, or to plus or minus infinity, the series is called oscillatory.
Proof of the Ratio Test
Let a series be \( u_1 + u_2 + u_3 + \ldots \). We assume that the inequalities in the statement hold.
Convergence Case
From the first part of the statement:
\( \dfrac{u_2}{u_1} < r \), \( \dfrac{u_3}{u_2} < r \), … where \( r < 1 \).
Therefore:
$$u_1 + u_2 + u_3 + \ldots = u_1 \left(1 + \frac{u_2}{u_1} + \frac{u_3}{u_1} + \ldots \right)$$
$$= u_1 \left(1 + \frac{u_2}{u_1} + \frac{u_3}{u_2} \cdot \frac{u_2}{u_1} + \ldots \right)$$
$$< u_1(1 + r + r^2 + \ldots)$$
Therefore, \( \sum u_n < u_1(1 + r + r^2 + \ldots) \)
$$\sum u_n < \lim_{n \to \infty} \frac{u_1(1 – r^n)}{1 – r}$$
Since \( r < 1 \), as \( n \to \infty \), \( r^n \to 0 \).
Therefore \( \sum u_n < \dfrac{u_1}{1-r} = k \), where \( k \) is a fixed number.
Hence \( \sum u_n \) is convergent. ∎
Divergence Case
Since \( \dfrac{u_{n+1}}{u_n} > 1 \), we have \( \dfrac{u_2}{u_1} > 1 \), \( \dfrac{u_3}{u_2} > 1 \), and so on.
Therefore:
\( u_2 > u_1 \)
\( u_3 > u_2 > u_1 \)
\( u_4 > u_3 > u_2 > u_1 \)
and so on.
Therefore \( \sum u_n = u_1 + u_2 + u_3 + \ldots + u_n > nu_1 \).
By taking \( n \) sufficiently large, \( nu_1 \) can be made greater than any fixed quantity.
Hence the series is divergent. ∎
Rigorous Proofs
From the statement of the theorem, it is necessary that \( \forall n: a_n \ne 0 \); otherwise \( \dfrac{a_{n+1}}{a_n} \) is not defined.
Here, \( \left| \dfrac{a_{n+1}}{a_n} \right| \) denotes either the absolute value of \( \dfrac{a_{n+1}}{a_n} \), or the complex modulus of \( \dfrac{a_{n+1}}{a_n} \).
Absolute Convergence
Suppose \( l < 1 \).
Let us take \( \epsilon > 0 \) such that \( l + \epsilon < 1 \).
Then: \( \exists N: \forall n > N: \left| \dfrac{a_n}{a_{n-1}} \right| < l + \epsilon \)
Thus:
$$|a_n| = \left| \frac{a_n}{a_{n-1}} \right| \left| \frac{a_{n-1}}{a_{n-2}} \right| \cdots \left| \frac{a_{N+2}}{a_{N+1}} \right| |a_{N+1}|$$
$$< (l + \epsilon)^{n-N-1} |a_{N+1}|$$
By the Sum of Infinite Geometric Progression, \( \displaystyle \sum_{n=1}^\infty (l + \epsilon)^n \) converges.
So by the corollary to the comparison test, it follows that \( \displaystyle \sum_{n=1}^\infty |a_n| \) converges absolutely. ∎
Divergence
Suppose \( l > 1 \).
Let us take \( \epsilon > 0 \) small enough that \( l – \epsilon > 1 \).
Then, for a sufficiently large \( N \), we have:
$$|a_n| = \left| \frac{a_n}{a_{n-1}} \right| \left| \frac{a_{n-1}}{a_{n-2}} \right| \cdots \left| \frac{a_{N+2}}{a_{N+1}} \right| |a_{N+1}|$$
$$> (l – \epsilon)^{n-N-1} |a_{N+1}|$$
But \( (l – \epsilon)^{n-N-1} |a_{N+1}| \to \infty \) as \( n \to \infty \).
So \( \displaystyle \sum_{n=1}^\infty a_n \) diverges. ∎
Important Remarks
- When \( \dfrac{u_{n+1}}{u_n} = 1 \), the test fails. You’ll need another test (like Raabe’s test or the root test).
- Alternative form: The series \( \sum u_n \) of positive terms is convergent if \( \displaystyle \lim_{n \to \infty} \dfrac{u_n}{u_{n+1}} > 1 \) and divergent if \( \displaystyle \lim_{n \to \infty} \dfrac{u_n}{u_{n+1}} < 1 \).
- This alternative form is often more convenient in practical applications since it avoids dealing with fractions of fractions.
Suggested Reading
Analysis I, Fourth Edition
Author: Terence Tao
Hardcover : 376 pages
Worked Example
Verify whether the infinite series \( \dfrac{x}{1 \cdot 2} + \dfrac{x^2}{2 \cdot 3} + \dfrac{x^3}{3 \cdot 4} + \ldots \) is convergent or divergent.
Solution
We have \( u_{n+1} = \dfrac{x^{n+1}}{(n+1)(n+2)} \) and \( u_n = \dfrac{x^n}{n(n+1)} \).
Therefore:
$$\lim_{n \to \infty} \frac{u_n}{u_{n+1}} = \lim_{n \to \infty} \left(1 + \frac{2}{n}\right) \frac{1}{x} = \frac{1}{x}$$
When \( 1/x > 1 \), i.e., \( x < 1 \), the series is convergent.
When \( x > 1 \), the series is divergent.
When \( x = 1 \), the ratio test fails (limit equals 1). We need another approach.
For \( x = 1 \): \( u_n = \dfrac{1}{n(n+1)} = \dfrac{1}{n^2} \left(1 + \frac{1}{n}\right)^{-1} \)
Expanding: \( u_n = \dfrac{1}{n^2} \left(1 – \frac{1}{n} + \frac{1}{n^2} – \ldots \right) \)
Take \( v_n = \dfrac{1}{n^2} \). Now \( \displaystyle \lim_{n \to \infty} \dfrac{u_n}{v_n} = 1 \), a non-zero finite quantity.
But \( \sum v_n = \sum \dfrac{1}{n^2} \) is convergent (p-series with \( p = 2 > 1 \)).
By the limit comparison test, \( \sum u_n \) is also convergent when \( x = 1 \).
Frequently Asked Questions
What is D’Alembert’s Ratio Test?
D’Alembert’s Ratio Test determines whether an infinite series converges or diverges by examining the limit of the ratio of consecutive terms. If the limit is less than 1, the series converges absolutely. If greater than 1, it diverges. If equal to 1, the test is inconclusive.
Who was Jean le Rond d’Alembert?
Jean le Rond d’Alembert (1717-1783) was a French mathematician, physicist, and philosopher. He made major contributions to mechanics, calculus, and the theory of differential equations. The ratio test is one of several mathematical concepts named after him, including d’Alembert’s principle in mechanics.
When does the ratio test fail?
The ratio test is inconclusive when the limit equals 1. In such cases, you need to use other tests like Raabe’s test, the root test, or comparison tests. Classic examples where the ratio test fails include the harmonic series (divergent) and the series 1/n² (convergent), both giving limit = 1.
What is absolute convergence?
A series Σaₙ converges absolutely if the series of absolute values Σ|aₙ| converges. Absolute convergence implies ordinary convergence, but not vice versa. The ratio test, when it gives a limit less than 1, proves absolute convergence, which is a stronger result than conditional convergence.
What is the difference between convergent and divergent series?
A convergent series has partial sums that approach a finite limit. A divergent series has partial sums that either grow without bound, oscillate, or fail to approach any limit. For example, 1 + 1/2 + 1/4 + … converges to 2, while 1 + 2 + 3 + … diverges to infinity.
How do you apply the ratio test step by step?
First, find the general term aₙ. Second, write aₙ₊₁ by replacing n with n+1. Third, form the ratio |aₙ₊₁/aₙ|. Fourth, compute the limit as n→∞. Finally, conclude: if limit 1, divergent; if limit = 1, use another test.
What is an oscillatory series?
An oscillatory series has partial sums that neither converge to a finite limit nor diverge to infinity. Instead, they bounce between values without settling. The classic example is 1 – 1 + 1 – 1 + …, whose partial sums alternate between 0 and 1 forever.
Why compare with geometric series in the proof?
The geometric series 1 + r + r² + … converges when |r| < 1 and equals 1/(1-r). In the ratio test proof, we show that terms of our series are bounded by terms of a convergent geometric series. By the comparison test, our series must also converge.
What is the alternative form of the ratio test?
Instead of computing lim(aₙ₊₁/aₙ), you can compute lim(aₙ/aₙ₊₁). If this limit is greater than 1, the series converges. If less than 1, it diverges. This form is sometimes more convenient because it avoids nested fractions in the calculation.
What other tests can you use when the ratio test fails?
When the ratio test gives limit = 1, try Raabe’s test, the root test (Cauchy), the integral test, or comparison tests (direct or limit comparison). For alternating series, the Leibniz test works. The choice depends on the series structure. Raabe’s test is specifically designed for cases where the ratio test fails.