What if the limit equals 1?

The ratio test is inconclusive. You must use another method, such as the limit comparison test, the integral test, or the alternating series test.

Can the ratio test be used for rational functions like $1/k^2$?

Yes, but the limit will always evaluate to $L = 1$, rendering the test inconclusive. Use the p-series test or limit comparison test for rational expressions.

How do you handle a negative term or alternating series?

The absolute value operation inside the limit $|a_{k+1} / a_k|$ removes any alternating signs or negative constant multipliers prior to taking the limit.

How to test a series for convergence with the ratio test

The ratio test evaluates the limit of the absolute value of the ratio of consecutive terms in an infinite series. It determines absolute convergence, divergence, or yields an inconclusive result. This test applies primarily to series containing factorials, exponentials, or products of polynomial and exponential functions where term cancellation simplifies the limit.

The setup

Identify the general term $a_k$ from the infinite series $\sum a_k$ . Substitute $k+1$ for every instance of $k$ to form the expression for the next term $a_{k+1}$ . Place both terms inside the absolute value ratio $|a_{k+1} / a_k|$ .

The steps

Construct the ratio $|a_{k+1} / a_k|$ and multiply by the reciprocal to clear complex fractions. 2. Simplify the algebraic expression by canceling factorials and combining exponents with identical bases. 3. Evaluate the limit $L = \lim_{k o \infty} |a_{k+1} / a_k|$ using standard limit techniques for rational functions. 4. Conclude based on $L$ : if $L < 1$ , the series converges absolutely; if $L > 1$ or $L = \infty$ , the series diverges; if $L = 1$ , the test is inconclusive.

Checking the result

Verify that absolute value bars were applied before evaluating the limit, removing any alternating components like $(-1)^k$ . Ensure limit evaluation accurately factored out the dominant terms of polynomials in the numerator and denominator.

Common errors

A frequent error is incorrectly expanding complex factorials. For example, replacing $k$ with $k+1$ in $(2k)!$ yields $(2(k+1))! = (2k+2)!$ , not $(2k+1)!$ . Another common mistake is concluding convergence when $L=1$ instead of applying a different convergence test.

Worked example

Test the series $\sum_{k=1}^{\infty} \frac{k^2}{3^k}$ for convergence using the ratio test.

Let $a_k = \frac{k^2}{3^k}$ . Form the next term: $a_{k+1} = \frac{(k+1)^2}{3^{k+1}}$ . Construct the absolute ratio: $|\frac{a_{k+1}}{a_k}| = |\frac{(k+1)^2}{3^{k+1}} \cdot \frac{3^k}{k^2}|$ . Since all terms are positive for $k \ge 1$ , drop the absolute values. Simplify the expression by grouping similar terms: $\frac{(k+1)^2}{k^2} \cdot \frac{3^k}{3^{k+1}} = (\frac{k+1}{k})^2 \cdot \frac{1}{3}$ . Take the limit as $k o \infty$ : $L = \lim_{k o \infty} (1 + \frac{1}{k})^2 \cdot \frac{1}{3} = (1)^2 \cdot \frac{1}{3} = \frac{1}{3}$ . Since $L = 1/3 < 1$ , the series converges absolutely.

FAQ

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Answer form

References: Stewart Calculus 8th Edition, Chapter 11.6 · OpenStax Calculus Volume 2, Chapter 5.6 · Thomas' Calculus 14th Edition, Chapter 10.5