10.8 - Ratio Test for Convergence

10.8 Ratio Test for Convergence

Prerequisites: Two Essential Skills

Factorial manipulation

n! = n \cdot (n-1) \cdot (n-2) \cdots 3 \cdot 2 \cdot 1

For example,

5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120

The key property:

(n+1)! = (n+1) \cdot n!

which lets you cancel in fractions:

\frac{(n+1)!}{n!} = \frac{(n+1) \cdot n!}{n!} = n+1

Similarly:

\frac{(n+2)!}{n!} = \frac{(n+2)(n+1) \cdot n!}{n!} = (n+2)(n+1)

Exponent manipulation

a^{m+n} = a^m \cdot a^n

For example:

\frac{5^{n+2}}{5^n} = \frac{5^n \cdot 5^2}{5^n} = 25

These two properties are what make the ratio test work.

Statement of the Ratio Test

For a series

\sum_{n=1}^{\infty} a_n

with positive terms, define:

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

\boxed{\begin{cases} L < 1 & \implies \text{Series CONVERGES} \\ L > 1 & \implies \text{Series DIVERGES} \\ L = 1 & \implies \text{INCONCLUSIVE (use another test)} \end{cases}}

The ratio test needs positive terms. It doesn’t apply directly to alternating series, though you can apply it to the absolute values.

When the Ratio Test Is Inconclusive

The ratio test gives you nothing when

L = 1

, and that happens constantly with p-series.

Example 1: the harmonic series (which we know diverges)

Test

\sum_{n=1}^{\infty} \frac{1}{n}

Here

a_n = \frac{1}{n}

, so:

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

Since

L = 1

, the ratio test is inconclusive. We already know the harmonic series diverges, but the ratio test can’t tell us that.

Example 2: p-series with $p = 2$ (which we know converges)

Test

\sum_{n=1}^{\infty} \frac{1}{n^2}

L = \lim_{n \to \infty} \frac{\frac{1}{(n+1)^2}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = \left(\lim_{n \to \infty} \frac{n}{n+1}\right)^2 = 1^2 = 1

Again inconclusive. We know it converges (p-series with

p = 2 > 1

), but the ratio test won’t confirm it.

General rule: don’t use the ratio test on p-series. It always gives

L = 1

When the Ratio Test Shows Convergence

Example 3: exponential and polynomial terms

Does

\sum_{n=1}^{\infty} \frac{n^2}{2^n}

converge or diverge?

Step 1: write out $a_n$ and $a_{n+1}$

a_n = \frac{n^2}{2^n} \qquad a_{n+1} = \frac{(n+1)^2}{2^{n+1}}

Step 2: form the ratio

L = \lim_{n \to \infty} \frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}} = \lim_{n \to \infty} \frac{(n+1)^2}{2^{n+1}} \cdot \frac{2^n}{n^2}

Step 3: simplify with exponent properties

= \lim_{n \to \infty} \frac{(n+1)^2}{n^2} \cdot \frac{2^n}{2^n \cdot 2} = \lim_{n \to \infty} \frac{(n+1)^2}{n^2} \cdot \frac{1}{2}

Step 4: evaluate

= \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^2 \cdot \frac{1}{2} = 1^2 \cdot \frac{1}{2} = \frac{1}{2}

Step 5: conclude. Since

L = \frac{1}{2} < 1

, the series converges by the ratio test.

Example 4: exponentials top and bottom

Does

\sum_{n=1}^{\infty} \frac{(n+1)^3 \cdot 4^{n+1}}{7^n}

converge or diverge?

Step 1: identify terms

a_n = \frac{(n+1)^3 \cdot 4^{n+1}}{7^n} \qquad a_{n+1} = \frac{(n+2)^3 \cdot 4^{n+2}}{7^{n+1}}

Step 2: form the ratio

L = \lim_{n \to \infty} \frac{(n+2)^3 \cdot 4^{n+2}}{7^{n+1}} \cdot \frac{7^n}{(n+1)^3 \cdot 4^{n+1}}

Step 3: simplify

= \lim_{n \to \infty} \frac{(n+2)^3}{(n+1)^3} \cdot \frac{4^{n+2}}{4^{n+1}} \cdot \frac{7^n}{7^{n+1}} = \lim_{n \to \infty} \frac{(n+2)^3}{(n+1)^3} \cdot 4 \cdot \frac{1}{7}

Step 4: evaluate

= \lim_{n \to \infty} \left(\frac{n+2}{n+1}\right)^3 \cdot \frac{4}{7} = 1^3 \cdot \frac{4}{7} = \frac{4}{7}

Step 5: conclude. Since

L = \frac{4}{7} < 1

, the series converges by the ratio test.

Series with Factorials

This is where the ratio test really earns its keep.

Example 5: factorial in the numerator

Does

\sum_{n=1}^{\infty} \frac{n!}{5^n}

converge or diverge?

Step 1: identify terms

a_n = \frac{n!}{5^n} \qquad a_{n+1} = \frac{(n+1)!}{5^{n+1}}

Step 2: form the ratio

L = \lim_{n \to \infty} \frac{(n+1)!}{5^{n+1}} \cdot \frac{5^n}{n!}

Step 3: simplify the factorial using

(n+1)! = (n+1) \cdot n!

= \lim_{n \to \infty} \frac{(n+1) \cdot n!}{n!} \cdot \frac{5^n}{5^n \cdot 5} = \lim_{n \to \infty} \frac{n+1}{5}

Step 4: evaluate

= \infty

Step 5: conclude. Since

L = \infty > 1

, the series diverges by the ratio test.

Example 6: factorial with exponential

Does

\sum_{n=1}^{\infty} \frac{3^n}{n!}

converge or diverge?

Step 1: identify terms

a_n = \frac{3^n}{n!} \qquad a_{n+1} = \frac{3^{n+1}}{(n+1)!}

Step 2: form the ratio

L = \lim_{n \to \infty} \frac{3^{n+1}}{(n+1)!} \cdot \frac{n!}{3^n}

Step 3: simplify

= \lim_{n \to \infty} \frac{3^n \cdot 3}{(n+1) \cdot n!} \cdot \frac{n!}{3^n} = \lim_{n \to \infty} \frac{3}{n+1}

Step 4: evaluate

= 0

Step 5: conclude. Since

L = 0 < 1

, the series converges by the ratio test.

Factorials grow faster than exponentials, so whenever a factorial sits in the denominator, convergence is likely.

Advanced Factorial Manipulation

Example 7: double factorial expression

Does

\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 \cdot 3^n}

converge or diverge?

Step 1: identify terms

a_n = \frac{(2n)!}{(n!)^2 \cdot 3^n} \qquad a_{n+1} = \frac{(2n+2)!}{[(n+1)!]^2 \cdot 3^{n+1}}

Step 2: expand the factorials

(2n+2)! = (2n+2)(2n+1) \cdot (2n)!

(n+1)! = (n+1) \cdot n! \quad \Rightarrow \quad [(n+1)!]^2 = (n+1)^2 \cdot (n!)^2

Step 3: substitute and simplify

L = \lim_{n \to \infty} \frac{(2n+2)(2n+1)(2n)!}{(n+1)^2(n!)^2 \cdot 3^{n+1}} \cdot \frac{(n!)^2 \cdot 3^n}{(2n)!} = \lim_{n \to \infty} \frac{(2n+2)(2n+1)}{(n+1)^2 \cdot 3}

Factor a 2 out of

(2n+2)

= \lim_{n \to \infty} \frac{2(n+1)(2n+1)}{3(n+1)^2} = \lim_{n \to \infty} \frac{2(2n+1)}{3(n+1)} = \lim_{n \to \infty} \frac{4n+2}{3n+3} = \frac{4}{3}

Step 4: conclude. Since

L = \frac{4}{3} > 1

, the series diverges by the ratio test.

Step-by-Step Strategy

Write out $a_n$ and $a_{n+1}$ explicitly

Form the ratio $\frac{a_{n+1}}{a_n}$ (multiply by the reciprocal)

Simplify:

Factor factorials: $(n+1)! = (n+1) \cdot n!$

Separate exponents: $a^{n+k} = a^n \cdot a^k$

Cancel common terms

Evaluate $L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$ (this is a horizontal asymptote question)

Conclude: $L < 1$ converges, $L > 1$ diverges, $L = 1$ is inconclusive

When to Use the Ratio Test

Best for: - Series with factorials (

n!

(2n)!

, and so on) - Series with exponentials (

a^n

e^n

) - Series with both

Not useful for: - p-series (always gives

L = 1

) - Simple rational functions (use comparison or the p-series test)

When the ratio test gives $L = 1$ , fall back on the nth term test, comparison tests, the integral test, or the p-series test.

Summary of Tests Learned

Test	Best For	Limitation
nth term	Quick divergence check	Can’t prove convergence
Geometric	$ar^n$ form	Only for geometric series
p-series	$\frac{1}{n^p}$	Only for p-series
Integral	Easy-to-integrate functions	Requires integration
Comparison	Bounded by a known series	Need a good comparison
Alternating	Alternating signs	Only for alternating series
Ratio	Factorials, exponentials	Inconclusive at $L = 1$

Key Takeaways

Ratio test formula: $L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$

Convergence: $L < 1$ . Divergence: $L > 1$ . Inconclusive: $L = 1$

Factorial property: $(n+1)! = (n+1) \cdot n!$ , which lets you cancel

Exponent property: $a^{n+k} = a^n \cdot a^k$ , same idea

Best for factorials and exponentials. It’s extremely powerful there

Always inconclusive for p-series. Don’t waste your time

Simplify carefully. Factor before you cancel

Evaluate the limit like a horizontal asymptote, comparing highest powers

Factorials grow faster than exponentials: $\frac{a^n}{n!}$ usually converges

Exponentials grow faster than polynomials: $\frac{n^k}{a^n}$ converges if $a > 1$

Write out $a_n$ and $a_{n+1}$ first. It prevents most errors

When $L = 1$ , switch tests. The ratio test has told you nothing

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