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Section 10.2 : More on Sequences

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

\[\left\{ {\frac{{4 - n}}{{2n + 3}}} \right\}_{n = 1}^\infty \]

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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 having the increasing/decreasing information will probably make the determining the bounds (if any exist) somewhat easier.
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For this problem let’s get the increasing/decreasing information first.

For Problems 1 – 3 in this section it was easy enough to just ask what happens if we increase \(n\) to determine the increasing/decreasing information for this problem. However, in this case, increasing \(n\) will increase both the numerator and denominator and so it would be somewhat more difficult to do that analysis here.

Therefore, we will resort to some quick Calculus I to determine increasing/decreasing information. We can define the following function and take its derivative.

\[f\left( x \right) = \frac{{4 - x}}{{2x + 3}}\hspace{0.5in} \Rightarrow \hspace{0.5in}f'\left( x \right) = \frac{{ - 11}}{{{{\left( {2x + 3} \right)}^2}}}\]

We can clearly see that the derivative will always be negative for \(x \ne - \frac{3}{2}\) and so the function is decreasing for \(x \ne - \frac{3}{2}\). Therefore, because the function values are the same as the sequence values when \(x\) is an integer we can see that the sequence, which starts at \(n = 1\), must also be decreasing and hence it is also monotonic.

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

First, because the sequence is decreasing we can see that the first term of the sequence will be the largest and hence will also be an upper bound for the sequence. So, the sequence is bounded above by \(\frac{3}{5}\) (i.e. the \(n = 1\) sequence term).

Next let’s look for the lower bound (if it exists). For this problem let’s first take a quick look at the limit of the sequence terms. In this case the limit of the sequence terms is,

\[\mathop {\lim }\limits_{n \to \infty } \frac{{4 - n}}{{2n + 3}} = - \frac{1}{2}\]

Recall what this limit tells us about the behavior of our sequence terms. The limit means that as \(n \to \infty \) the sequence terms must be getting closer and closer to \( - \frac{1}{2}\).

Now, for a second, let’s suppose that that \( - \frac{1}{2}\) is not a lower bound for the sequence terms and let’s also keep in mind that we’ve already determined that the sequence is decreasing (means that each successive term must be smaller than (i.e. below) the previous one…).

So, if \( - \frac{1}{2}\) is not a lower bound then we know that somewhere there must be sequence terms below (or smaller than) \( - \frac{1}{2}\). However, because we also know that terms must be getting closer and closer to \( - \frac{1}{2}\) and we’ve now assumed there are terms below \( - \frac{1}{2}\) the only way for that to happen at this point is for at least a few sequence terms to increase up towards \( - \frac{1}{2}\) (remember we’ve assumed there are terms below this!). That can’t happen however because we know the sequence is a decreasing sequence.

Okay, what was the point of all this? Well recall that we got to this apparent contradiction to the decreasing nature of the sequence by first assuming that \( - \frac{1}{2}\) was not a lower bound. Since making this assumption led us to something that can’t possibly be true that in turn means that \( - \frac{1}{2}\) must in fact be a lower bound since we’ve shown that sequence terms simply cannot go below this value!

Therefore, the sequence is bounded below by \( - \frac{1}{2}\).

Finally, because this sequence is both bounded above and bounded below the sequence is bounded.

Before leaving this problem a quick word of caution. The limit of a sequence is not guaranteed to be a bound (upper or lower) for a sequence. It will only be a bound under certain circumstances and so we can’t simply compute the limit and assume it will be a bound for every sequence! Can you see a condition that will allow the limit to be a bound?