Ratio Test

Last Updated : 23 Jul, 2025

Ratio Test is a mathematical tool used to determine whether an infinite series converges or diverges. It's particularly useful when dealing with series that involve factorials, exponentials, or more complex terms where other tests might be cumbersome or inconclusive.

Ratio Test is a method used in calculus to determine the convergence or divergence of an infinite series. It involves calculating the limit of the ratio of consecutive terms in the series.

  • The Ratio Test is particularly effective for series with terms that grow or shrink rapidly, such as those involving powers or factorials.
  • If the limit L < 1, the series converges absolutely (Convergence means the sum of a series reaches a specific value as more terms are added.)
  • If L > 1, the series diverges (Divergence means the sum of a series grows without bound as more terms are added.)
  • If L = 1, the test does not provide a definitive answer and other methods must be used.

Example 1:

\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots

The common ratio is \frac{1}{2}​. Since \frac{1}{2}​ is less than 1, this series converges.

Example 2:

1 + 2 + 4 + 8 + 16 + \cdots

The common ratio is 2. Since 2 is more than 1, this series diverges.

Here's how Ratio test works:

Given a series ∑a_n ​, compute the limit:

L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

Where a_n​ represents the nth term of the series.

Then,

  • If L < 1, the series converges.
  • If L > 1, the series diverges.
  • If L = 1, the test is inconclusive (meaning the series could either converge or diverge, and another test might be needed).

How the Ratio Test Works

The Ratio Test evaluates the convergence or divergence of a series \sum_{n=1}^{\infty} a_n by analyzing the limit of the ratio of consecutive terms.

Step 1: Identify the terms of the series.

Consider the general term a_n of the series. For instance, in the series \sum_{n=1}^{\infty} \frac{n!}{n^n}, the general term is a_n = \frac{n!}{n^n}.

Step 2: Calculate the ratio of consecutive terms.

Compute the ratio \frac{a_{n+1}}{a_n}. This ratio will often simplify and reveal how the terms of the series behave as n becomes large.

Step 3: Take the limit as n approaches infinity.

Find the limit L as n \to \infty of the absolute value of this ratio:

L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

Step 4: Interpret the result.

  • If L < 1, the series converges absolutely. This means that not only does the series converge, but it does so even if we consider the absolute values of its terms.
  • If L > 1 or L = \infty, the series diverges.
  • If L = 1, the test is inconclusive. In this case, the Ratio Test does not provide information about the convergence of the series, and other tests must be employed.

Examples

Convergent because L < 1

Consider the series ∑_{n=1}^{\infty} \frac{3^n}{n!}

Applying the ratio test, one computes the limit

L = \lim_{n \to \infty} \left|\frac{3^{n+1}}{{(n+1)}!}\frac{n!}{3^n} \right|= \lim_{n \to \infty} \frac{3}{(n+1)}=0 <1

Thus the series converges.

Divergent because L > 1

Consider the series ∑_{n=1}^{\infty} \frac{n!}{3^n}

Applying the ratio test, one computes the limit

L = \lim_{n \to \infty} \left|\frac{(n+1)!}{3^{n+1}} \frac{3^n}{n!} \right|= \lim_{n \to \infty} \frac{(n+1)}{3}=∞ >1

Thus the series diverges.

Inconclusive because L = 1

Consider the series ∑_{n=1}^{\infty} {n!}

Applying the ratio test, one computes the limit

L = \lim_{n \to \infty} \left|\frac{(n+1)!}{n!} \right|= \lim_{n \to \infty} {(1+\frac {1}{n!})}=1

Thus the series inconclusive can not solve by ratio test

When Doesn't the Ratio Test Work?

The Ratio Test may not work or be conclusive in the following situations:

  • Inconclusive Result
    • The Ratio Test is inconclusive when the limit L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|= 1.
    • In this case, the test does not determine whether the series converges or diverges. You would need to apply a different convergence test, such as the Root Test, Comparison Test, or Integral Test, to reach a conclusion.
  • Series with Terms that Don't Simplify Well
    • The Ratio Test may be difficult to apply or less effective when the terms of the series don't lead to a straightforward or simple ratio, such as in some oscillating series or series with complex expressions.

In summary, the Ratio Test doesn't work when the limit of the ratio equals 1, leading to an inconclusive result, or when the series terms are not conducive to simplification through the test. In these cases, alternative convergence tests should be considered.

series can only be conditionally convergent when the Ratio Test is inconclusive

  • If the Ratio Test shows L < 1, it confirms that the series converges absolutely (which is a stronger condition than conditional convergence).
  • If L > 1, it confirms that the series diverges.
  • However, if L = 1, the Ratio Test doesn't provide any information, and this is where conditional convergence can occur. For example, the alternating harmonic series converges conditionally, but the Ratio Test doesn't help because its limit is 1.

Therefore, a series can only be conditionally convergent when the Ratio Test is inconclusive (i.e., L = 1), meaning that the convergence behavior is subtle and cannot be determined solely by the Ratio Test.

Practice Problems on Ratio Test

To deepen your understanding of the Ratio Test, try solving the following problems:

Problem 1: Determine the convergence of the series \sum_{n=1}^{\infty} \frac{n!}{n^n} using the Ratio Test.

Problem 2: Apply the Ratio Test to the series \sum_{n=1}^{\infty} \frac{e^n}{n!} and determine its convergence.

Problem 3: Check the convergence of the series \sum_{n=1}^{\infty} \frac{2^n}{n^2} using the Ratio Test.

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