5.6 Ratio and Root Tests

Big idea

The comparison tests work by measuring a series against a known one. The two tests in this section work differently: they measure a series against itself. Instead of finding a benchmark, you ask how each term compares to the one before it, or how the term compares to an $n$th power.

The motivation is the geometric series, the one family whose behavior is completely understood. In a geometric series the ratio of consecutive terms is exactly $r$ at every step, and the series converges precisely when $|r| < 1$. The ratio test takes that observation and relaxes it: if the ratio of consecutive terms is not constant but settles on a limit $\rho$, then for large $n$ the series is eventually behaving like a geometric series with ratio $\rho$, and the same threshold applies. The root test does the same from the other side, extracting $r$ from $r^n$ by taking an $n$th root.

Both tests conclude absolute convergence rather than plain convergence, since both work with $|a_n|$ throughout. That is a strength, because absolute convergence is the stronger statement and it makes the tests usable on series with any pattern of signs at all. It is also what makes these the tests of choice for power series in the next chapter, where the goal is to find the values of a variable for which a series converges absolutely.

There is one hard limitation and it is not an accident of the proof. When the limit is exactly $1$, both tests are silent, and the silence is unavoidable: convergent and divergent series both produce the value $1$. Since every p-series produces $1$, these tests cannot decide a single p-series, and any series whose terms are a ratio of polynomials is the wrong customer for them. Recognizing that in advance saves you from computing a limit that was never going to help.

Decoder

If the sizes of consecutive terms have a limiting ratio less than one, the series converges absolutely; if that ratio exceeds one, the series diverges; and if it equals one, the test decides nothing.

The word “limiting” matters. Individual ratios may be above $1$ for a while; only the limit is consulted. And the word “absolutely” matters: the ratio is taken between absolute values, so the conclusion in the convergent case is the stronger one, and signs play no role in the computation.

The divergent case deserves a note on why it is even possible, given that no other test extracts divergence from a comparison of this kind. When $\rho > 1$, the sizes of the terms are eventually increasing, so they cannot approach $0$, and the divergence test finishes the job. Divergence here is really the divergence test in disguise, which is also why a limit of exactly $1$ cannot be pushed into that argument.

The classic mistake is treating $\rho = 1$ as a verdict. It is not a near miss or a hint; it is an instruction to use a different test. A second, more mechanical mistake is mishandling factorials. The expression $\frac{(n+1)!}{n!}$ equals $n+1$, not $n! \cdot$ anything you need to expand, and writing $(n+1)! = (n+1)\,n!$ before dividing makes every one of these problems short.

Definitions and results

The ratio test. For a series $\sum a_n$ with nonzero terms, let

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

If $\rho < 1$ the series converges absolutely. If $\rho > 1$, or the ratio grows without bound, the series diverges. If $\rho = 1$ the test is inconclusive.

The root test. For a series $\sum a_n$, let

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

with the same three conclusions: absolute convergence for $\rho < 1$, divergence for $\rho > 1$, no information for $\rho = 1$.

Why the threshold is one. If $\rho < 1$, pick any $c$ with $\rho < c < 1$. Beyond some index the sizes satisfy $|a_{n+1}| \le c|a_n|$, so the tail is dominated termwise by a convergent geometric series with ratio $c$, and direct comparison finishes it.

Which test suits which series. Use the ratio test when factorials or products that telescope under division appear. Use the root test when the whole term is raised to the $n$th power, since the root removes the exponent cleanly. When both apply they give the same value of $\rho$.

Useful ratios. $\frac{(n+1)!}{n!} = n+1$, and $\frac{r^{\,n+1}}{r^{\,n}} = r$, and $\frac{(n+1)^{p}}{n^{p}} = \left(1 + \frac1n\right)^{p} \to 1$.

Useful roots. $\sqrt[n]{r^{\,n}} = |r|$, and $\sqrt[n]{n} \to 1$, and $\sqrt[n]{c} \to 1$ for any constant $c > 0$.

Where both tests fail. Any series whose terms are a ratio of polynomials gives $\rho = 1$ under both tests. Since $\sum \frac1n$ diverges and $\sum \frac{1}{n^2}$ converges and both produce $\rho = 1$, no test that sees only $\rho$ can separate them. Send such series to the comparison or integral tests instead.

Relationship to the other tests. A ratio or root verdict of convergence is a statement about $\sum |a_n|$, so it also settles the original series. A verdict of divergence comes from the terms failing to approach zero, so it is stronger than mere failure of a comparison.

Worked examples

A factorial in the denominator

Decide $\displaystyle\sum_{n=1}^{\infty}\frac{3^{\,n}}{n!}$.

The ratio test is built for this. Write out the ratio and cancel:

$$ \left|\frac{a_{n+1}}{a_n}\right| = \frac{3^{\,n+1}}{(n+1)!}\cdot\frac{n!}{3^{\,n}} = \frac{3 \cdot 3^{\,n}}{(n+1)\,n!}\cdot\frac{n!}{3^{\,n}} = \frac{3}{n+1} $$

The limit is $\rho = 0 < 1$, so the series converges absolutely. The same computation with any constant in place of $3$ gives $\rho = 0$, which is the precise sense in which factorials beat exponentials.

A factorial against a power with a matching exponent

Decide $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^{\,n}}$.

Both parts grow enormously, so the verdict is not obvious. Form the ratio:

$$ \left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{(n+1)^{\,n+1}}\cdot\frac{n^{\,n}}{n!} = \frac{(n+1)\,n^{\,n}}{(n+1)^{\,n+1}} = \frac{n^{\,n}}{(n+1)^{\,n}} = \frac{1}{\left(1 + \frac1n\right)^{n}} $$

The denominator tends to $e$, so $\rho = \frac1e \approx 0.368 < 1$ and the series converges absolutely. Note that $\rho$ is a genuine number strictly between $0$ and $1$ here, so the series converges at roughly a geometric rate with ratio about $0.37$, which is useful information the verdict alone would not give.

An entire term raised to the nth power

Decide $\displaystyle\sum_{n=1}^{\infty}\left(\frac{2n+1}{5n+3}\right)^{\,n}$.

A ratio here would be unpleasant, but the root test removes the exponent in one step:

$$ \sqrt[n]{|a_n|} = \frac{2n+1}{5n+3} \longrightarrow \frac{2}{5} $$

So $\rho = \frac25 < 1$ and the series converges absolutely. The intuition matches: for large $n$ the base is close to $\frac25$, so the terms behave like $\left(\frac25\right)^n$, a convergent geometric series.

Two series the tests cannot tell apart

Apply the ratio test to $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^2+1}$ and to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$.

For the first,

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

since numerator and denominator are both cubics with leading coefficient $1$. For the second,

$$ \left|\frac{a_{n+1}}{a_n}\right| = \frac{n^2}{(n+1)^2} = \frac{1}{\left(1 + \frac1n\right)^{2}} \longrightarrow 1 $$

Both give $\rho = 1$, and the two series have different verdicts: limit comparison of $\frac{n}{n^2+1}$ with $\frac1n$ gives the ratio $\frac{n^2}{n^2+1} \to 1$, so the first diverges alongside the harmonic series, while the second is a convergent p-series with $p = 2$. The ratio test was never going to distinguish them, and the right response to $\rho = 1$ is to switch tests immediately rather than to compute more carefully.

Practice

One skill covering both tests. Expect factorials, powers of $n$, terms raised to the $n$th power, and series designed so that one of the two tests is far less work than the other. When the limit comes out as $1$, say so and name a test that would settle the question.

Quiz

Five items on computing the ratio and root limits, reading the three cases correctly, and recognizing series for which both tests are inconclusive.