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### Section 10.2 : More on Sequences

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

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

\[f\left( x \right) = \frac{{2{x^2} - 1}}{x}\hspace{0.5in} \Rightarrow \hspace{0.5in}f'\left( x \right) = \frac{{2{x^2} + 1}}{{{x^2}}}\]We can clearly see that the derivative will always be positive for \(x \ne 0\) and so the function is increasing for \(x \ne 0\). 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 \(n = 2\), must also be **increasing** and hence it is also **monotonic**.

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

First, it is hopefully obvious that all the terms are positive for our range of \(n \ge 2\) and so the sequence is **bounded below by zero**. We could also use the fact that the sequence is increasing the first term would have to be the smallest term in the sequence and so a better lower bound would be the first sequence term which is \(\frac{7}{2}\). Either would work for this problem.

Now let’s see what we can determine about an upper bound (provided it has one of course…).

We know that the function is increasing but that doesn’t mean there is no upper bound. Take a look at Problems 1 and 3 above. Each of those were decreasing sequences and yet they had a lower bound. Do not make the mistake of assuming that an increasing sequence will not have an upper bound or a decreasing sequence will not have a lower bound. Sometimes they will and sometimes they won’t.

For this sequence we’ll need to approach any potential upper bound a little differently than the previous problems. Let’s first compute the following limit of the terms,

\[\mathop {\lim }\limits_{n \to \infty } \frac{{2{n^2} - 1}}{n} = \mathop {\lim }\limits_{n \to \infty } \left( {2n - \frac{1}{n}} \right) = \infty \]Since the limit of the terms is infinity we can see that the terms will increase without bound. Therefore, in this case, there really is **no upper bound** for this sequence. Please remember the warning above however! Just because this increasing sequence had no upper bound does not mean that every increasing sequence will not have an upper bound.

Finally, because this sequence is bounded below but not bounded above the sequence is **not bounded**.