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,

|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|

(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:

|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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