Proof of the Ratio Test

From Consecutive Ratios to a Series Test

The Ratio Test was stated in the previous tutorial. Here we assemble its proof from the geometric comparison and necessary-condition results established earlier. The key point is that a ratio limit strictly below one can be replaced, eventually, by a fixed factor below one. A ratio limit strictly above one instead forces the terms away from zero.

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.

Proof. First suppose $L<1$. The Proposition (A Ratio Limit Below One Gives Geometric Domination) says that for some $q$ with $L<q<1$, there is an index $N$ such that \[ |a_{n+1}|\leq q|a_n| \qquad (n\geq N). \] The Geometric Domination Bound then gives convergence of $\sum_{n=N}^{\infty}|a_n|$. Adding the finitely many terms before $N$ preserves convergence, so $\sum_{n=1}^{\infty}|a_n|$ converges. Since $\sum_{n=1}^{\infty}|a_n|$ converges, the original series converges absolutely by definition, and hence converges by the theorem on absolute convergence proved in a later tutorial.

Now suppose $1<L\leq+\infty$. The Proposition (A Ratio Limit Above One Forces Divergence) says that $a_n$ does not tend to zero. The Necessary Condition for Series Convergence requires the terms of every convergent series to tend to zero. Therefore $\sum_{n=1}^{\infty}a_n$ diverges. These two cases establish the asserted conclusions. When $L=1$, neither proposition applies; examples below show why no conclusion is available from the ratio limit alone. $\square$

This proof is a chain of implications rather than a separate estimate for every term. In the convergence case, the ratio limit supplies eventual geometric domination, and the geometric comparison controls the entire tail. In the divergence case, the ratio limit prevents the terms from satisfying a necessary condition for convergence. The assumption that terms are eventually nonzero ensures that the ratios used in both arguments are defined.

A More Precise Description of the Convergent Tail

The geometric tail bound gives a useful upper estimate, but positive terms with a ratio limit below one have a sharper asymptotic description: far out in the series, the tail is approximately a fixed multiple of its first term. The multiplier is the sum of a geometric series with ratio equal to the limiting consecutive-term ratio.

Theorem (Asymptotic Geometric Tail): Suppose $a_n>0$ for all $n\geq N_0$ and $$ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L,\qquad 0\leq L<1. $$ For $M\geq N_0$, define $T_M=\sum_{n=M}^{\infty}a_n$. Then $$ \lim_{M\to\infty}\frac{T_M}{a_M}=\frac{1}{1-L}. $$

Proof. Choose $q$ such that $L<q<1$. By the definition of the ratio limit, there is an index $N\geq N_0$ for which $a_{n+1}/a_n\leq q$ whenever $n\geq N$. Fix $M\geq N$. For every integer $j\geq0$, set \[ u_{M,j}=\frac{a_{M+j}}{a_M}. \] In particular, $u_{M,0}=1$. For $j\geq1$, writing the quotient as a product gives \[ u_{M,j}=\prod_{i=0}^{j-1}\frac{a_{M+i+1}}{a_{M+i}}. \] Each factor is at most $q$, so $0<u_{M,j}\leq q^j$. For each fixed $j$, every factor in this finite product tends to $L$ as $M\to\infty$; hence $u_{M,j}\to L^j$. This also holds for $j=0$, since both sides equal one.

The terms are positive and the geometric bound makes their tail summable. Thus \[ \frac{T_M}{a_M}=\sum_{j=0}^{\infty}u_{M,j}. \] We show that these sums tend to $\sum_{j=0}^{\infty}L^j$. Given $\varepsilon>0$, choose an integer $J$ large enough that \[ \sum_{j=J+1}^{\infty}q^j=\frac{q^{J+1}}{1-q}<\frac{\varepsilon}{3}. \] For every $M\geq N$, the tail $\sum_{j=J+1}^{\infty}u_{M,j}$ is at most this quantity. Also $0\leq L\leq q$, so the tail $\sum_{j=J+1}^{\infty}L^j$ is at most $q^{J+1}/(1-q)<\varepsilon/3$. For the finite sum from $j=0$ to $J$, the termwise limits $u_{M,j}\to L^j$ imply that, for all sufficiently large $M$, \[ \left|\sum_{j=0}^{J}u_{M,j}-\sum_{j=0}^{J}L^j\right|<\frac{\varepsilon}{3}. \] Combining the finite-sum difference and the two tails shows \[ \left|\frac{T_M}{a_M}-\sum_{j=0}^{\infty}L^j\right|<\varepsilon. \] Since $0\leq L<1$, the geometric series sums to $1/(1-L)$. This proves the stated limit. $\square$

Unlike the Ratio Test itself, this result describes the relative size of the remainder, not merely whether the series converges. Its positivity hypothesis matters: it allows the tail to be compared directly with the first term without cancellation.

Worked Applications

Worked Example: A Signed Series with a Polynomial Factor

Consider the series with terms $a_n=(-1)^n(n^2+1)/4^n$, for $n\geq1$. The absolute values are positive, and their consecutive-term ratio is

\frac{|a_{n+1}|}{|a_n|} =\frac{((n+1)^2+1)/4^{n+1}}{(n^2+1)/4^n} =\frac{n^2+2n+2}{4(n^2+1)}.

Dividing numerator and denominator by $n^2$ shows that this ratio tends to $1/4$. Since $1/4<1$, the Ratio Test proves that the series converges absolutely. The signs alternate, but they do not affect the ratio test because it uses absolute values.

Worked Example: A Ratio Limit Above One

Let $a_n=5^n/(n+1)$, for $n\geq1$. Then

\frac{|a_{n+1}|}{|a_n|} =\frac{5^{n+1}/(n+2)}{5^n/(n+1)} =5\frac{n+1}{n+2} \longrightarrow 5.

Because $5>1$, the divergence case of the Ratio Test applies. In particular, the terms cannot tend to zero: eventually they increase by at least a fixed factor greater than one. The series diverges by the Necessary Condition for Series Convergence.

Worked Example: A Tail Compared with Its First Term

Take $a_n=2^{-n}(1+1/n)$, for $n\geq1$. These terms are positive, and

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

The Ratio Test gives convergence. The Asymptotic Geometric Tail Theorem gives more: if $T_M=\sum_{n=M}^{\infty}a_n$, then

\frac{T_M}{a_M}\longrightarrow\frac{1}{1-1/2}=2.

Thus, for large $M$, the tail is approximately $2a_M$. This is an asymptotic statement: it does not say that $T_M=2a_M$ exactly for any particular $M$.

Worked Example: Ratio Limit One at Two Different Outcomes

For integers $n\geq2$, consider first $a_n=1/(n\log n)$. Its ratio is

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

The series diverges. Indeed, the Integral Test applies to $f(x)=1/(x\log x)$ for $x\geq2$, since $f$ is positive, continuous, and decreasing there, and \[ \int_2^b\frac{dx}{x\log x}=\log(\log b)-\log(\log 2)\longrightarrow+\infty. \] Now take $b_n=1/(n(\log n)^2)$. Its ratio is

\frac{b_{n+1}}{b_n} =\frac{n}{n+1}\left(\frac{\log n}{\log(n+1)}\right)^2 \longrightarrow1.

But this series converges: the Integral Test applies, and the substitution $u=\log x$ gives \[ \int_2^{\infty}\frac{dx}{x(\log x)^2} =\int_{\log 2}^{\infty}\frac{du}{u^2} =\frac{1}{\log 2}<\infty. \] Both ratio limits equal one, even though the series have different outcomes. The Ratio Test cannot decide either series from that limit.

What the Proof Does—and Does Not—Tell Us

When the limit is below one, the choice of $q$ creates a margin between the limiting ratio and one. That margin is what makes geometric comparison possible. When the limit is above one, a fixed factor greater than one eventually bounds the ratios from below, so the terms fail to tend to zero. The proof needs only eventual behavior; changing finitely many initial terms does not affect convergence.

A ratio limit of one contains too little information to decide convergence. The logarithmic examples show that even terms with closely related forms can decay at different enough rates to produce opposite outcomes. The Asymptotic Geometric Tail Theorem also has a narrower scope than the Ratio Test: it requires positive terms, whereas the Ratio Test’s convergence conclusion applies to signed terms through absolute convergence.

Takeaway: The Ratio Test follows by converting a strict ratio limit into an eventual geometric bound or an eventual growth bound. For positive terms with ratio limit below one, the tail divided by its first term tends to the corresponding geometric-series sum.

Check Your Understanding

Use the proof and the tail result to answer the following questions.

Which earlier results complete the convergence case of the Ratio Test, and which complete the divergence case?
Why does the proof for $L<1$ choose a number $q$ strictly between $L$ and one?
In the Asymptotic Geometric Tail Theorem, why is it necessary to control the terms of the normalized tail uniformly in $M$?
If positive terms have consecutive-term ratio tending to $2/5$, what is the limit of the tail divided by its first term?
Why can two series whose positive-term ratios both tend to one have different convergence behavior?

What You'll Learn