The Ratio Test

Why Compare Consecutive Terms?

The comparison tests and Cauchy condensation use information about terms across a range of indices. The Ratio Test takes a different approach: it asks how each term compares with the one immediately before it. If the absolute values shrink by roughly the same factor at every step, the terms eventually behave like those of a geometric series.

For a real series $\sum_{n=1}^{\infty}a_n$, the ratio to examine is $|a_{n+1}|/|a_n|$. Absolute values matter because the test first determines whether $\sum |a_n|$ converges. We require the terms to be nonzero from some point onward, so that these ratios are defined. A finite number of zero terms or other initial terms does not affect convergence, by the earlier result on finite changes to a series.

Theorem (Ratio Test): Suppose $(a_n)$ is a real sequence that is nonzero for all sufficiently large $n$, and suppose $$ \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}=L, $$ where $L$ may be a nonnegative real number or $+\infty$. If $L<1$, then $\sum_{n=1}^{\infty}a_n$ converges absolutely. If $1<L\leq+\infty$, then $\sum_{n=1}^{\infty}a_n$ diverges. If $L=1$, the test gives no conclusion.

The threshold is one. A ratio below one eventually forces repeated decay; a ratio above one eventually forces terms to grow rather than tend to zero. At the threshold, however, ratios alone do not reveal whether the terms decay quickly enough for the series to converge. The results below make the geometric comparison precise and give a useful tail estimate.

Geometric Domination and a Tail Bound

Theorem (Geometric Domination Bound): Suppose $N\geq1$, $0\leq q<1$, and a sequence $(a_n)$ satisfies $$ |a_{n+1}|\leq q|a_n| \qquad\text{for every }n\geq N. $$ Then $\sum_{n=N}^{\infty}|a_n|$ converges, and for every $M\geq N$, $$ \sum_{n=M}^{\infty}|a_n|\leq\frac{|a_M|}{1-q}. $$

Proof. Repeatedly applying the assumed inequality gives, for every integer $j\geq0$,

|a_{M+j}|\leq q^j|a_M|.

For $j=0$, this is equality. If it holds for some $j$, then \[ |a_{M+j+1}|\leq q|a_{M+j}|\leq q^{j+1}|a_M|, \] so the claim follows for all $j$ by induction. Therefore, for every $K\geq M$,

\sum_{n=M}^{K}|a_n| \leq |a_M|\sum_{j=0}^{K-M}q^j \leq\frac{|a_M|}{1-q}.

The last inequality follows from the finite geometric-sum formula and $0\leq q<1$. These partial sums are nondecreasing, since their terms are nonnegative, and are bounded above by $|a_M|/(1-q)$. By the earlier theorem on bounded increasing partial sums, they converge. Their limit satisfies the same bound. This proves both convergence and the stated tail estimate. $\square$

The estimate applies starting at any $M\geq N$, not just at the first index where the ratio bound holds. It is therefore useful for controlling remainders: once a geometric bound has been established, the size of the tail is bounded in terms of its first term.

Proposition (A Ratio Limit Below One Gives Geometric Domination): Suppose $a_n\neq0$ for all $n\geq N_0$, and $$ \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}=L<1. $$ For any $q$ satisfying $L<q<1$, there is an integer $N\geq N_0$ such that $|a_{n+1}|\leq q|a_n|$ for every $n\geq N$. Consequently, the absolute-value tail converges and satisfies the bounds in the Geometric Domination Bound.

Proof. Set $\varepsilon=q-L$, which is positive. By the definition of the limit, there is an integer $N\geq N_0$ such that whenever $n\geq N$,

\left|\frac{|a_{n+1}|}{|a_n|}-L\right|<\varepsilon.

In particular, \[ \frac{|a_{n+1}|}{|a_n|}<L+\varepsilon=q. \] Because $|a_n|>0$ on this range, multiplication by $|a_n|$ gives $|a_{n+1}|\leq q|a_n|$. The Geometric Domination Bound now applies. $\square$

Together, these results explain the convergence side of the Ratio Test: a ratio limit strictly below one can be replaced, eventually, by a fixed ratio $q$ that is still below one. The terms are then bounded by a geometric sequence, and the tail estimate quantifies the resulting control.

What Happens Above One?

Proposition (A Ratio Limit Above One Forces Divergence): Suppose $a_n\neq0$ for all sufficiently large $n$, and $$ \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}=L>1, $$ where $L$ may be $+\infty$. Then $a_n$ does not tend to zero. Therefore $\sum_{n=1}^{\infty}a_n$ diverges.

Proof. Choose a real number $q$ with $1<q<L$ if $L$ is finite. If $L=+\infty$, choose any $q>1$. By the definition of the limit (or divergence to $+\infty$), there is an $N$ such that \[ \frac{|a_{n+1}|}{|a_n|}>q \] for every $n\geq N$. Thus $|a_{n+1}|>q|a_n|$, and repeated application gives

|a_{N+k}|>q^k|a_N| \qquad (k\geq1).

Since $|a_N|>0$ and $q>1$, the right-hand side does not tend to zero; indeed, it grows without bound. Hence $a_n$ cannot tend to zero. The Necessary Condition for Series Convergence says that the terms of a convergent series must tend to zero, so the series diverges. $\square$

This direction is stronger than merely saying that the terms fail to be small: their absolute values eventually increase by at least a fixed factor greater than one. The conclusion uses the necessary condition for convergence, not a comparison with a divergent positive series.

Worked Applications

Worked Example: A Factorial Series

Consider $\sum_{n=0}^{\infty}3^n/n!$, where $0!=1$. Its terms are positive, and the ratio of consecutive terms is

\frac{3^{n+1}/(n+1)!}{3^n/n!}=\frac{3}{n+1}.

As $n$ tends to infinity, this ratio tends to zero, which is less than one. The Ratio Test therefore gives absolute convergence. It also gives a concrete tail estimate: for every $n\geq5$, \[ \frac{3}{n+1}\leq\frac{3}{6}=\frac12. \] Writing $a_n=3^n/n!$, the Geometric Domination Bound with $q=1/2$ shows that for every $M\geq5$,

\sum_{n=M}^{\infty}\frac{3^n}{n!} \leq 2\frac{3^M}{M!}.

The ratio limit establishes convergence; the eventual bound by $1/2$ supplies an explicit estimate for the remainder.

Worked Example: Polynomial Growth with Exponential Decay

Consider $\sum_{n=1}^{\infty}n^2/5^n$. The ratio of consecutive terms is

\frac{(n+1)^2/5^{n+1}}{n^2/5^n} =\frac{(n+1)^2}{5n^2} =\frac{1}{5}\left(1+\frac{1}{n}\right)^2.

Since $1+1/n$ tends to one, the ratio tends to $1/5$, which is below one. Thus the series converges absolutely. For a direct geometric estimate, when $n\geq1$, $1+1/n\leq2$, so the ratio is at most $4/5$. Taking $a_n=n^2/5^n$, the tail bound gives, for $M\geq1$,

\sum_{n=M}^{\infty}\frac{n^2}{5^n} \leq\frac{a_M}{1-4/5} =5\frac{M^2}{5^M}.

The polynomial factor changes the individual terms, but its effect on the consecutive-term ratio tends to one. The exponential factor leaves the limiting ratio $1/5$, which determines the outcome.

Worked Example: The Boundary Case Can Converge

Let $a_n=1/n^2$ for $n\geq1$. Then

\frac{a_{n+1}}{a_n} =\frac{1/(n+1)^2}{1/n^2} =\left(\frac{n}{n+1}\right)^2 \longrightarrow1.

The Ratio Test gives no conclusion. Nevertheless, the series $\sum_{n=1}^{\infty}1/n^2$ converges by the p-Series Convergence Criterion, since $p=2>1$. This example shows that a ratio limit of one is not a criterion for divergence.

Worked Example: The Same Boundary Case Can Diverge

Now let $a_n=1/n$. Its consecutive-term ratio is

\frac{a_{n+1}}{a_n} =\frac{1/(n+1)}{1/n} =\frac{n}{n+1} \longrightarrow1.

Again, the Ratio Test gives no conclusion. In this case, the series $\sum_{n=1}^{\infty}1/n$ diverges, as follows from the p-Series Convergence Criterion with $p=1$. The two examples have the same ratio limit but different outcomes. A limit of one does not distinguish them.

Interpreting the Test Carefully

For a ratio limit below one, the test establishes absolute convergence, even if the original terms have alternating signs or a more complicated sign pattern. Indeed, for $m>n$, $\left|\sum_{k=n+1}^{m}a_k\right|\leq\sum_{k=n+1}^{m}|a_k|$; since $\sum |a_k|$ converges, these tails tend to zero, so the partial sums of $\sum a_k$ are Cauchy and the series converges. For a ratio limit above one, the test establishes divergence because the terms fail the necessary condition for convergence.

The boundary case is a common source of overinterpretation. A ratio approaching one says that consecutive absolute values become close relative to their size. It does not specify how quickly those values themselves decay. The examples $1/n$ and $1/n^2$ illustrate the distinction: in both cases consecutive terms have ratio tending to one, but one series diverges and the other converges.

The ratio limit must also be computed from consecutive terms, with care about indices and signs. For a series with potentially negative terms, use $|a_{n+1}|/|a_n|$, not $a_{n+1}/a_n$. If some terms vanish repeatedly, the ratio may not be defined; the stated test applies when the terms are eventually nonzero. Initial terms do not decide convergence, but the eventual hypotheses do matter.

Takeaway: A ratio limit strictly below one gives eventual geometric decay and absolute convergence; a limit strictly above one prevents the terms from tending to zero. At a limit of one, use another method: the Ratio Test alone cannot decide convergence.

Check Your Understanding

Use the ratio criterion and the geometric tail estimate to answer the following questions.

Why does a limit $L<1$ allow the consecutive-term ratios to be bounded by some fixed $q<1$ eventually?
If $|a_{n+1}|\leq(2/3)|a_n|$ for every $n\geq N$, what bound does the Geometric Domination Bound give for $\sum_{n=M}^{\infty}|a_n|$, where $M\geq N$?
Why does a ratio limit greater than one imply divergence of the series?
What does the Ratio Test conclude for $\sum 1/n^3$, and why is its conclusion limited?
Give the ratio limit for the terms $n^2/5^n$, and explain how the polynomial factor enters the calculation.

What You'll Learn