# Proving that a particular sequence is a Cauchy sequence

Here is one of my favorite homework problems from my Advanced Calculus (introductory real analysis) class at Western Michigan University. It is problem 7 from Chapter 1.6 of Advanced Calculus: Theory and Practice by John Petrovic.

Let $$0 < r < 1$$ and let $$M>0$$. And suppose that $$\{a_n\}$$ is a sequence such that, $$\forall n \in \mathbb{N}$$, $$|a_{n+1} – a_n| \leq Mr^n$$. Prove that $$\{a_n\}$$ is a Cauchy sequence.

Proof: Recall that $$\{a_n\}$$ is a Cauchy sequence i.f.f. $$\forall \epsilon > 0$$, $$\exists N$$ such that, $$\forall m > n > N$$, $$|a_m – a_n| < \epsilon$$. Informally, the given information about $$\{a_n\}$$ means that any two consecutive terms of the sequence are separated by no more than a number $$M$$ multiplied by a number $$r^n$$ that is less than $$1$$; how much less than $$1$$ depends on how large $$n$$ is. So the product $$Mr^n$$ approaches $$0$$ as $$n$$ approaches $$\infty$$. We have to prove that a sequence with this property also has the defining property of a Cauchy sequence. If you're not fresh up on Cauchy sequences, one important thing about them is that, since $$m>n$$, there might be several (or millions or quintillions) of terms between $$a_n$$ and $$a_m$$, which we represent by $$a_{n+1}, a_{n+2}, … , a_{m-1}$$. Thus, using the ol’ trick of adding and subtracting the same thing (many times) so as not to change the value of the expression,

$$\begin{eqnarray} |a_m – a_n| &=& |(a_m – a_{m-1}) + (a_{m-1} – a_{m-2}) + … + (a_{n+2} – a_{n+1}) + (a_{n+1} – a_n)| \nonumber \\[5pt] &\leq& |a_m – a_{m-1}| + |a_{m-1} – a_{m-2}| + … + |a_{n+2} – a_{n+1}| + |a_{n+1} – a_n| \end{eqnarray}$$

(by the Triangle Inequality).

But notice: Since we are given that $$|a_{n+1} – a_n| \leq Mr^n$$ for all $$n \in N$$, it follows that $$|a_m – a_{m-1}| \leq Mr^{m-1}$$ and $$|a_{m-1} – a_{m-2}| \leq Mr^{m-2}$$ and $$|a_{n+2} – a_{n+1}| \leq Mr^{n+1}$$ and $$|a_{n+1} – a_n)| \leq Mr^n$$. Therefore, we can pick up where we left off:

$$\begin{eqnarray} |a_m – a_{m-1}| + … + |a_{n+1} – a_n| &\leq& |Mr^{m-1}| + |Mr^{m-2}| + … + |Mr^{n+1}| + |Mr^n| \nonumber \\[5pt] &=& Mr^n (r^{m-1-n} + r^{m-2-n} + … + r + 1) \nonumber \\[5pt] &<& Mr^n \left(\frac{1}{1-r}\right) ~ [by ~ a ~ property ~ of ~ geometric ~ series] \nonumber \\[5pt] &=& \frac{Mr^n}{1-r} \nonumber \\[5pt] \end{eqnarray}$$ We have shown that $$|a_m - a_n| < \frac{Mr^n}{1-r}$$. To prove that $$\{a_n\}$$ is a Cauchy sequence, we must show that $$|a_m - a_n| < \epsilon$$. Thus, we should choose $$N$$ such that $$\frac{Mr^N}{1-r} < \epsilon$$, and it will follow that the same inequality holds for $$n$$, which is greater than $$N$$. Observe: $$\begin{eqnarray} \frac{Mr^N}{1-r} &<& \epsilon \nonumber \\[5pt] r^N &<& \frac{\epsilon(1-r)}{M} \nonumber \\[5pt] \ln(r^N) &<& \ln\left(\frac{\epsilon(1-r)}{M}\right) \nonumber \\[5pt] N\ln r &<& \ln\left(\frac{\epsilon(1-r)}{M}\right) \nonumber \\[5pt] N &>& \frac{\ln\left(\frac{\epsilon(1-r)}{M}\right)}{\ln r} ~ [switch ~ inequalities ~ because ~ \ln r < 0!] \end{eqnarray}$$ Then for all $$m > n > N$$, $$|a_m – a_n| < \epsilon$$ for any $$\epsilon > 0$$, meaning $$\{a_n\}$$ is a Cauchy sequence. $$\blacksquare$$

I wasn’t able to include a hyperlink in an equation display, but the property of geometric series I referred to above, to justify a strict inequality, can be found here, for example.

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