# 8.7: Ratio Test

• Bob Dumas and John E. McCarthy
• University of Washington and Washington University in St. Louis

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One of the uses of the order-completeness of the real numbers is proving that an infinite sequence converges, without having to know much about the number to which it converges. In Chapter 5 we allude to the ratio test in claiming that the Taylor polynomial for the exponential function evaluated at a real number $$a, \sum_{k=0}^{\infty} \frac{a^{k}}{k !}$$, converges. How do we prove that an infinite sum converges? If we have an idea of its limit, we might show that the sequence of partial sums approaches this value. This is how we prove that the geometric sum with ratio less than 1 converges. Many important mathematical functions are defined by infinite sums, and the limit of the sum defines the value of the function. In this case we need to show that the sum converges using properties of the real numbers.

DEFINITION. Absolute convergence Let $$\left\langle a_{n}\right\rangle$$ be an infinite sequence. If the infinite sum $\sum_{k=0}^{\infty}\left|a_{k}\right|$ converges then the infinite $$\operatorname{sum} \sum_{k=0}^{\infty} a_{k}$$ is said to converge absolutely.

LEMMA 8.8. If an infinite sum converges absolutely, then it converges.

Proof. Assume $$\sum_{k=0}^{\infty} a_{k}$$ converges absolutely. We show that the sequence of partial sums of this series, $$\left\langle s_{n} \mid n \in \mathbb{N}\right\rangle$$, is a Cauchy sequence. For $$n \in \mathbb{N}$$, let $b_{n}=\left|a_{n}\right| .$ Then $$\sum_{k=0}^{\infty} b_{k}$$ converges. Let $$\left\langle t_{n} \mid n \in \mathbb{N}\right\rangle$$ be the sequence of partial sums of $$\sum_{k=0}^{\infty} b_{k}$$. By Theorem $$8.7,\left\langle t_{n}\right\rangle$$ is a Cauchy sequence. Let $$\varepsilon>0$$. Then there is $$N \in \mathbb{N}$$ such that for any $$n \geq m \geq N$$, $\left|t_{n}-t_{m}\right| \leq \varepsilon .$ By a generalization of the triangle inequality (see Exercise 8.24) $\left|s_{n}-s_{m}\right|=\left|\sum_{k=m+1}^{n} a_{k}\right| \leq \sum_{k=m+1}^{n} b_{k}=\left|t_{n}-t_{m}\right|<\varepsilon$ Hence $$\left\langle s_{n}\right\rangle$$ is a Cauchy sequence and converges. Therefore $$\sum_{k=0}^{\infty} a_{k}$$ converges.

THEOREM 8.9. Ratio test Suppose $$\left\langle a_{k}\right\rangle$$ is an infinite sequence of real numbers and that there is $$N \in \mathbb{N}$$ and a positive real number $$r<1$$ such that for all $$n \geq N$$, $\left|\frac{a_{n+1}}{a_{n}}\right| \leq r .$ Then $$\sum_{k=0}^{\infty} a_{k}$$ converges.

Proof. Let $$\sum_{k=0}^{\infty} a_{k}$$ be an infinite sum with terms satisfying the hypothesis. For $$n \in \mathbb{N}$$, let $$b_{n}=\left|a_{n}\right|$$. By assumption, there is $$N \in \mathbb{N}$$ and a positive real number $$r<1$$ such that for all $$n \geq N$$, $\frac{b_{n+1}}{b_{n}} \leq r .$ We may assume without loss of generality that $$N=0$$, since the series $$\sum_{k=0}^{\infty} b_{k}$$ converges iff $$\sum_{k=N}^{\infty} b_{k}$$ converges, and if necessary we may ignore finitely many terms of the infinite sum. We claim that for all $$n \in \mathbb{N}$$, $b_{n} \leq b_{0} r^{n} .$ If $$n=0$$ the claim is obvious. Assume the claim holds at $$n$$. By assumption, $\frac{b_{n+1}}{b_{n}} \leq r .$ Therefore $b_{n+1} \leq r b_{n} \leq r b_{0} r^{n} \leq b_{0} r^{n+1} .$ By Exercise 5.28, the geometric sum with radius $$-1<r<1$$ converges to $$\frac{1}{1-r}$$. Therefore, for any $$n \in \mathbb{N}$$, $s_{n}:=\sum_{k=0}^{n} b_{k} \leq \sum_{k=0}^{n} b_{0} r^{k}=b_{0}\left(\sum_{k=0}^{\infty} r^{k}\right) \leq \frac{b_{0}}{1-r} .$ The sequence of partial sums, $$\left\langle s_{n}\right\rangle$$, is a monotonic bounded sequence and by Lemma $$8.5$$, converges. Therefore $$\sum_{k=0}^{\infty} a_{k}$$ converges absolutely. By Lemma $$8.8$$ the sum converges.

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