Paul's Online Notes
Paul's Online Notes
Home / Calculus II / Series & Sequences / More on Sequences
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 10.2 : More on Sequences

3. Determine if the following sequence is increasing, decreasing, not monotonic, bounded below, bounded above and/or bounded.

\[\left\{ {{3^{ - \,n}}} \right\}_{n = 0}^\infty \]

Show All Steps Hide All Steps

Hint : There is no one set process for finding all this information. Sometimes it is easier to find one set of information before the other and at other times it doesn’t matter which set of information you find first. For this sequence it might be a little easier to find the bounds (if any exist) if you first have the increasing/decreasing information.
Start Solution

For this problem let’s get the increasing/decreasing information first as that seems to be pretty simple and will help at least a little bit with the bounded information.

We’ all agree that, for our range of \(n \ge 0\), we have,

\[n < n + 1\]

This in turn gives,

\[{3^{ - n}} = \frac{1}{{{3^n}}} > \frac{1}{{{3^{n + 1}}}} = {3^{ - \left( {n + 1} \right)}}\]

So, if we define \({a_n} = {3^{ - n}}\) we have \({a_n} > {a_{n + 1}}\) for all \(n \ge 0\) and so the sequence is decreasing and hence is also monotonic.

Show Step 2

Now let’s see what bounded information we can get.

First, it is hopefully obvious that all the terms are positive and so the sequence is bounded below by zero.

Next, we saw in the first step that the sequence was decreasing and so the first term will be the largest term and so the sequence is bounded above by \({3^{ - \left( 0 \right)}} = 1\) (i.e. the \(n = 0\) sequence term).

Therefore, because this sequence is bounded below and bounded above the sequence is bounded.