Section 10.2 : More on Sequences
6. Determine if the following sequence is increasing, decreasing, not monotonic, bounded below, bounded above and/or bounded.
\[\left\{ {\frac{{ - n}}{{{n^2} + 25}}} \right\}_{n = 2}^\infty \]Show All Steps Hide All Steps
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.
\[f\left( x \right) = \frac{{ - x}}{{{x^2} + 25}}\hspace{0.5in} \Rightarrow \hspace{0.5in}f'\left( x \right) = \frac{{{x^2} - 25}}{{{{\left( {{x^2} + 25} \right)}^2}}}\]Hopefully, it’s fairly clear that the critical points of the function are \(x = \pm 5\). We’ll leave it to you to draw a quick number line or sign chart to verify that the function will be decreasing in the range \(2 \le x < 5\) and increasing in the range \(x > 5\). Note that we just looked at the ranges of \(x\) that correspond to the ranges of \(n\) for our sequence here.
Now, 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 \(n = 2\), has terms that increase and terms that decrease and hence the sequence is not an increasing sequence and the sequence is not a decreasing sequence. That also means that the sequence is not monotonic.
Show Step 2Now let’s see what bounded information we can get.
In this case, unlike many of the previous problems in this section, we don’t have a monotonic sequence. However, we can still use the increasing/decreasing information above to help us out with the bounds.
First, we know that the sequence is decreasing in the range \(2 \le n < 5\) and increasing in the range \(n > 5\). From our Calculus I knowledge we know that this means \(n = 5\) must be a minimum of the sequence terms and hence the sequence is bounded below by \(\frac{{ - 5}}{{50}} = - \frac{1}{{10}}\) (i.e. the \(n = 5\) sequence term).
Next let’s look for the upper 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,
\[\mathop {\lim }\limits_{n \to \infty } \frac{{ - n}}{{{n^2} + 25}} = 0\]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 zero.
Now, for a second, let’s look at just the portion of the sequence with \(n > 5\) and let’s further suppose that zero is not an upper bound for the sequence terms with \(n > 5\). Let’s also keep in mind that we’ve already determined that the sequence is increasing for \(n > 5\) (means that each successive term must be larger than (i.e. above) the previous one…).
So, if zero is not an upper bound (for \(n > 5\) ) then we know that somewhere there must be sequence terms with \(n > 5\) above (or larger than) zero. So, we know that terms must be getting closer and closer to zero and we’ve now assumed there are terms above zero. Therefore the only way for the terms to approach the limit of zero is for at least a few sequence terms with \(n > 5\) to decrease down towards zero (remember we’ve assumed there are terms above this!). That can’t happen however because we know that for \(n > 5\) the sequence is increasing.
Okay, what was the point of all this? Well recall that we got to this apparent contradiction to the increasing nature of the sequence for \(n > 5\) by first assuming that zero was not an upper bound for the portion of the sequence with \(n > 5\). Since making this assumption led us to something that can’t possibly be true that in turn means that zero must in fact be an upper bound for the portion of the sequence with \(n > 5\) since we’ve shown that sequence terms simply cannot go above this value!
Note that we’ve not yet actually shown that zero in an upper bound for the sequence and in fact it might not actually be an upper bound. However, what we have shown is that it is an upper bound for the vast majority of the sequence, i.e. for the portion of the sequence with \(n > 5\).
All we need to do to finish the upper bound portion of this problem off is check what the first few terms of the sequence are doing. There are several ways to do this. One is to just compute the remaining initial terms of the sequence to see if they are above or below zero. For this sequence that isn’t too bad as there are only 4 terms (\(n = 2,3,4,5\)). However, if there’d been several hundred terms that wouldn’t be so easy so let’s take a look at another approach that will always be easy to do in this case because we have the increasing/decreasing information for this initial portion of the sequence.
Let’s simply note that for the first part of this sequence we’ve already shown that the sequence is decreasing. Therefore, the very first sequence term of \( - \frac{2}{{29}}\) (i.e. the \(n = 2\) sequence term) will be the largest term for this initial bit of the sequence that is decreasing. This term is clearly less than zero and so zero will also be larger than all the remaining terms in the initial decreasing portion of the sequence and hence the sequence is bounded above by zero.
Finally, because this sequence is both bounded above and bounded below the sequence is bounded.
Before leaving this problem a couple of quick words of caution.
First, the limit of a sequence is not guaranteed to be a bound (upper or lower) for a sequence so be careful to not just always assume that the limit is an upper/lower bound for a sequence.
Second, as this problem has shown determining the bounds of a sequence can sometimes be a fairly involved process that involves quite a bit of work and lots of various pieces of knowledge about the other behavior of the sequence.