Ratio and Root Tests

Tests for repeated multiplication

Comparison with $1/n^p$ is effective when powers of $n$ control a series. Factorials and exponentials follow a different long-run structure: multiplying from one index to the next reveals more than comparing polynomial sizes. The ratio and root tests detect whether a series’ terms eventually shrink like the terms of a convergent geometric series. Both tests are commonly applied to absolute values, so they directly establish absolute convergence when successful.

For a series $\sum a_n$, define

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

The ratio test says that the series converges absolutely if $L<1$ and diverges if $L>1$ or if the ratio tends to infinity. When $L=1$, the test is inconclusive. The boundary is exactly what geometric intuition suggests: a ratio eventually less than one means that terms are controlled by a decaying geometric pattern, while a ratio greater than one prevents the terms from approaching zero.

Factorials simplify under a ratio

Consider

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

The terms may look small at first, but the ratio test exposes the factorial’s growth:

$$\left|\frac{a_{n+1}}{a_n}\right| =\frac{(n+1)!/5^{n+1}}{n!/5^n} =\frac{n+1}{5}.$$

This ratio tends to infinity, so the terms eventually grow rather than shrink to zero; the series diverges. Notice the cancellation: $(n+1)!=(n+1)n!$, and $5^{n+1}=5\cdot5^n$. Writing those expansions before canceling makes the ratio test systematic instead of magical.

Now compare

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

The ratio is $5/(n+1)$, which tends to zero. Therefore the series converges absolutely. Swapping numerator and denominator reversed the competition between exponential and factorial growth. The test reveals that $n!$ eventually dominates every fixed-base exponential, a fact that would be hard to see from a short list of decimal terms.

The conclusion is about the entire series, but it begins with a claim about eventual term behavior. If the ratio is eventually, say, less than $1/2$, then each later magnitude is at most half the preceding magnitude. Repeating that estimate bounds the tail by a geometric series. The formal ratio test packages this reasoning into a limit condition. It does not require every early ratio to be less than one, only that the long-run ratio be bounded by some number strictly below one.

When applying the test, parentheses protect the indexing. Write $a_{n+1}$ in full before forming the quotient, especially if the term includes $n!$, $(n+1)!$, or a power such as $c^n$. A common error is to replace $n$ by $n+1$ in only one factor. Checking a first numeric ratio separately can help catch this, but the algebraic substitution is the proof.

The root test reads per-index exponential scale

The root test uses

$$L=\lim_{n\to\infty}\sqrt[n]{|a_n|}.$$

It has the same conclusions: $L<1$ gives absolute convergence, $L>1$ gives divergence, and $L=1$ is inconclusive. Taking an $n$th root is especially useful when the term itself contains an expression raised to the $n$th power. It extracts the effective repeated multiplier without requiring a messy quotient of consecutive terms.

For example,

$$\sum_{n=1}^{\infty}\left(\frac{2n+1}{5n-3}\right)^n$$

has root-test expression

$$\sqrt[n]{\left|\left(\frac{2n+1}{5n-3}\right)^n\right|} =\left|\frac{2n+1}{5n-3}\right|\to\frac25.$$

Since $2/5<1$, the series converges absolutely. The root test sees that the base approaches a ratio smaller than one, which acts like a decaying geometric factor. A direct comparison could be possible, but the root test makes the exponential structure immediate.

The value one is not a verdict

The ratio test returns $L=1$ for both $\sum1/n$ and $\sum1/n^2$. The first diverges and the second converges. The root test also returns $1$ for these examples. Thus an inconclusive result is information about the test’s reach, not evidence for either outcome. Switch to a p-series comparison, integral test, or another method that matches the polynomial structure.

Similarly, do not use the ratio test merely because it is available. For a rational function of $n$ with no factorials or exponentials, the ratio commonly tends to one and wastes effort. A useful triage question is: does taking $a_{n+1}/a_n$ cause substantial cancellation, especially of factorials or powers? If yes, the ratio test is promising. If the term is visibly an $n$th power, the root test may be shorter.

Absolute values also deserve an explicit step. A negative base, alternating sign, or complex-looking factor can change signs from one term to the next, but the ratio and root tests classify the magnitudes. After absolute convergence is proved, ordinary convergence follows automatically. If a test establishes divergence because $L>1$, the terms’ magnitudes fail to approach zero; no cancellation pattern can repair that failure. This is why the divergence conclusion is just as decisive for an alternating-looking expression as it is for a positive series.

Retrieval check

Apply the ratio test to $\sum_{n=1}^{\infty}(n^3)/(4^n)$ and the root test to $\sum_{n=1}^{\infty}[3n/(7n+1)]^n$. In each calculation, write the absolute value, simplify the ratio or root before taking its limit, and state whether the conclusion is absolute convergence, divergence, or inconclusive. Then explain why the ratio test gives no conclusion for $\sum1/n^4$ even though that series converges. Unit VI now uses convergence tools to build functions from power series and to control their approximation error.