D’Alembert’s Ratio Test for Convergence of Series

Q: How do you apply the ratio test step by step?

Write down the general term an. Replace every n with n+1 to get an+1. Form the ratio |an+1/an| and simplify. Compute the limit as n → ∞. If the limit is less than 1, the series converges. If greater than 1, it diverges. If equal to 1, try another test. The key is to simplify the ratio before taking the limit.

Q: 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.

Q: Why does the proof compare with geometric series?

The geometric series 1 + r + r² + ... converges when |r| < 1 and equals 1/(1−r). In the convergence proof, we show that the terms of our series are bounded above by the terms of a convergent geometric series with ratio r < 1. By the comparison test, our series must also converge. The geometric series acts as a known 'benchmark' that we compare against.

Q: What other tests can you use when the ratio test is inconclusive?

When the ratio test gives limit = 1, try Raabe's test first since it's specifically built for this situation. The root test (Cauchy) is a good alternative, especially for terms of the form f(n)n. The integral test works when your general term comes from a positive, decreasing function. Comparison tests (direct or limit comparison) are useful when you can identify a simpler series with known behavior. For alternating series, the Leibniz test is usually the right tool.

Q: 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 and d'Alembert's formula for the wave equation.

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. It works best on series involving factorials, exponentials, and powers, where other tests either get messy or don’t apply at all.

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.

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 form avoids nested fractions, which makes the algebra cleaner in many problems. You’ll see this used in the worked examples below.

My free Foundation of Analysis book covers the D’Alembert and convergence tests for series, with more proper examples and more. Get it to your email inbox now.

How to Apply the Ratio Test

The ratio test follows the same five steps every time. Once you’ve done it a few times, it becomes mechanical. Here’s the process you should follow for every problem.

Step-by-Step Method

Write down the general term $ a_n $. Identify the pattern in the series and express the $ n $th term as a formula.
Write $ a_{n+1} $. Replace every $ n $ in your general term with $ n+1 $.
Form the ratio $ \left| \dfrac{a_{n+1}}{a_n} \right| $. Simplify as much as possible before taking the limit.
Compute $ \displaystyle \lim_{n \to \infty} \left| \dfrac{a_{n+1}}{a_n} \right| $. This is where most of the algebra happens.
Compare the limit to 1. If $ l < 1 $: convergent. If $ l > 1 $: divergent. If $ l = 1 $: the ratio test can’t help you, try another test.

The ratio test works best when the general term contains factorials (like $ n! $), exponentials (like $ 2^n $), or products of both. If your series only has polynomial terms (like $ 1/n^2 $), the limit will almost always equal 1, and you’ll need a different test.

Worked Examples

The best way to understand the ratio test is to work through problems of increasing difficulty. I’ve arranged these from straightforward to tricky, including a case where the test fails and you need a backup plan.

Example 1: Exponential Series (Divergence)

Test the series $ \displaystyle \sum_{n=1}^{\infty} \frac{2^n}{n} $ for convergence.

Here $ a_n = \dfrac{2^n}{n} $ and $ a_{n+1} = \dfrac{2^{n+1}}{n+1} $.

$$\left| \frac{a_{n+1}}{a_n} \right| = \frac{2^{n+1}}{n+1} \cdot \frac{n}{2^n} = \frac{2n}{n+1}$$

$$\lim_{n \to \infty} \frac{2n}{n+1} = \lim_{n \to \infty} \frac{2}{1 + 1/n} = 2 > 1$$

Since the limit is 2 (greater than 1), the series diverges. The exponential $ 2^n $ grows much faster than the linear $ n $ in the denominator.

Example 2: Factorial Series (Convergence)

Test the series $ \displaystyle \sum_{n=1}^{\infty} \frac{n!}{n^n} $ for convergence.

This is the kind of series where the ratio test really shines. Here $ a_n = \dfrac{n!}{n^n} $ and $ a_{n+1} = \dfrac{(n+1)!}{(n+1)^{n+1}} $.

$$\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!}$$

Since $ (n+1)! = (n+1) \cdot n! $, this simplifies to:

$$= \frac{(n+1) \cdot n!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n$$

$$= \left(\frac{1}{1 + 1/n}\right)^n$$

$$\lim_{n \to \infty} \left(\frac{1}{1 + 1/n}\right)^n = \frac{1}{e} \approx 0.368$$

Since $ 1/e < 1 $, the series converges. Notice how naturally the number $ e $ appears here. This happens often when factorials and powers of $ n $ compete in the same term.

Example 3: Parameter-Dependent Series (Mixed Result)

Test the series $ \dfrac{x}{1 \cdot 2} + \dfrac{x^2}{2 \cdot 3} + \dfrac{x^3}{3 \cdot 4} + \ldots $ for convergence.

We have $ u_{n+1} = \dfrac{x^{n+1}}{(n+1)(n+2)} $ and $ u_n = \dfrac{x^n}{n(n+1)} $.

Using the alternative form of the ratio test:

$$\lim_{n \to \infty} \frac{u_n}{u_{n+1}} = \lim_{n \to \infty} \left(1 + \frac{2}{n}\right) \frac{1}{x} = \frac{1}{x}$$

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 $.

This example is important because it shows the full workflow: apply the ratio test first, then switch to a different test when the ratio test is inconclusive. You’ll see this pattern repeatedly in exam problems.

Example 4: The Harmonic Series (Inconclusive Case)

Test the harmonic series $ \displaystyle \sum_{n=1}^{\infty} \frac{1}{n} $ using the ratio test.

Here $ a_n = \dfrac{1}{n} $ and $ a_{n+1} = \dfrac{1}{n+1} $.

$$\left| \frac{a_{n+1}}{a_n} \right| = \frac{n}{n+1}$$

$$\lim_{n \to \infty} \frac{n}{n+1} = 1$$

The ratio test is inconclusive. We know the harmonic series diverges (by the integral test or p-series test with $ p = 1 $), but the ratio test can’t tell us that. The same limit of 1 shows up for $ \sum 1/n^2 $, which converges. This is exactly why $ l = 1 $ means “try something else,” not “maybe it converges.”

Common Mistakes to Avoid

I’ve seen students lose marks on these mistakes more than anything else related to the ratio test. Most of them are easy to fix once you know what to watch for.

Forgetting absolute values. The test requires $ \left| \dfrac{a_{n+1}}{a_n} \right| $, not just $ \dfrac{a_{n+1}}{a_n} $. For alternating series, skipping the absolute value gives you a negative limit, which makes no sense in this context. Always use $ |a_{n+1}/a_n| $.

Concluding from $ l = 1 $. When the limit equals 1, the test tells you nothing. It doesn’t mean “the series might converge.” It means the ratio test cannot determine convergence or divergence. Both the harmonic series ($ \sum 1/n $, divergent) and the Basel series ($ \sum 1/n^2 $, convergent) give $ l = 1 $. You must switch to another test.

Using it on purely polynomial series. If your general term is a rational function of $ n $ (like $ n^2/(3n^3+1) $), the ratio test will almost always give $ l = 1 $. Don’t waste time. Use the comparison test or limit comparison test instead.

Simplification errors with factorials. Remember that $ (n+1)! = (n+1) \cdot n! $. A common mistake is expanding both factorials fully instead of canceling. The whole point of the ratio test is that consecutive terms cancel nicely.

When the Ratio Test Fails

The ratio test is inconclusive when $ l = 1 $. This happens more often than you’d expect, especially with series involving only polynomial terms. When it fails, you need a backup plan. Here are the tests to try, roughly in order of usefulness.

Raabe’s test is specifically designed for series where the ratio test gives $ l = 1 $. It looks at the rate at which the ratio approaches 1. If $ \displaystyle \lim_{n \to \infty} n\left(\frac{a_n}{a_{n+1}} – 1\right) > 1 $, the series converges. Raabe’s test picks up exactly where the ratio test leaves off.

The root test (Cauchy’s test) uses $ \lim_{n \to \infty} |a_n|^{1/n} $ instead of the ratio. It’s sometimes more powerful, particularly for series where the general term has the form $ f(n)^n $.

Comparison tests (direct or limit comparison) work well when you can identify a simpler series that behaves similarly. The limit comparison test was used in Example 3 above to handle the $ x = 1 $ case.

The integral test works when $ a_n = f(n) $ for some positive, decreasing, continuous function $ f $. It converts the convergence question into an improper integral.

For alternating series, the Leibniz test (alternating series test) is usually the right tool. The ratio test works on absolute convergence, so it may miss conditional convergence entirely.

Proof of the Ratio Test

The proof comes in two flavors: a direct argument using geometric series comparison, and a more rigorous version using epsilon-delta reasoning. Both reach the same conclusion, but exam expectations vary. If your course uses Rudin or Bartle, you’ll want the rigorous version. If you’re in a standard calculus course, the first proof is enough.

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:

$$u_1 + u_2 + u_3 + \ldots = u_1 \left(1 + \frac{u_2}{u_1} + \frac{u_3}{u_1} + \ldots \right)$$

$$= u_1 \left(1 + \frac{u_2}{u_1} + \frac{u_3}{u_2} \cdot \frac{u_2}{u_1} + \ldots \right)$$

$$< u_1(1 + r + r^2 + \ldots)$$

The right side is a geometric series with ratio $ r < 1 $, so it converges to $ \dfrac{u_1}{1-r} $. Since every partial sum of $ \sum u_n $ is bounded above by a fixed number, the series is convergent. ∎

Divergence Case

Since $ \dfrac{u_{n+1}}{u_n} > 1 $, each term is larger than the previous one:

$ 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 $ exceeds any fixed quantity. 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. ∎

Key Definitions

If you’re comfortable with what convergence and divergence mean, skip this section. These definitions are here for reference, particularly if your exam requires formal epsilon-based 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. The classic example is $ 1 – 1 + 1 – 1 + \ldots $, whose partial sums alternate between 0 and 1 forever.

Which Resources to Refer?

If you want to study more about this, you can refer to the top analysis books like Rudin, Tao etc.

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Principles of Mathematical Analysis (3rd Edition) by Walter Rudin

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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 and you need a different test like Raabe’s test or the root test.

When does the ratio test fail?

The ratio test is inconclusive when the limit equals 1. This typically happens with series whose general terms are purely polynomial (like 1/n or 1/n²). The harmonic series (divergent) and the series 1/n² (convergent) both give limit = 1, which is why l = 1 means ‘inconclusive,’ not ‘maybe.’ Switch to Raabe’s test, the root test, or comparison tests in these cases.

How do you apply the ratio test step by step?

Write down the general term a_n. Replace every n with n+1 to get a_n+1. Form the ratio |a_n+1/a_n| and simplify. Compute the limit as n → ∞. If the limit is less than 1, the series converges. If greater than 1, it diverges. If equal to 1, try another test. The key is to simplify the ratio before taking the limit.

What is the alternative form of the ratio test?

Instead of computing lim(a_n+1/a_n), you can compute lim(a_n/a_n+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.

Why does the proof compare with geometric series?

The geometric series 1 + r + r² + … converges when |r| < 1 and equals 1/(1−r). In the convergence proof, we show that the terms of our series are bounded above by the terms of a convergent geometric series with ratio r < 1. By the comparison test, our series must also converge. The geometric series acts as a known 'benchmark' that we compare against.

What other tests can you use when the ratio test is inconclusive?

When the ratio test gives limit = 1, try Raabe’s test first since it’s specifically built for this situation. The root test (Cauchy) is a good alternative, especially for terms of the form f(n)ⁿ. The integral test works when your general term comes from a positive, decreasing function. Comparison tests (direct or limit comparison) are useful when you can identify a simpler series with known behavior. For alternating series, the Leibniz test is usually the right tool.

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 and d’Alembert’s formula for the wave equation.