Calculus at Moravian University

1. \(\ds a_1=\frac{3^1}{1!} = 3;\quad\) \(\ds a_2= \frac{3^2}{2!} = \frac92;\quad\) \(\ds a_3 = \frac{3^3}{3!} = \frac92;\quad\) \(\ds a_4 = \frac{3^4}{4!} = \frac{27}8\quad\)

   We can plot the terms of a sequence with a scatter plot. The horizontal axis is used for the values of \(n\text{,}\) and the values of the terms are plotted on the vertical axis. To visualize this sequence, see Figure 9.1.4.(a).

2. \(a_1= 4+(-1)^1 = 3;\quad\) \(a_2 = 4+(-1)^2 = 5;\quad\) \(a_3=4+(-1)^3 = 3;\quad\) \(a_4 = 4+(-1)^4 = 5\quad\text{.}\)

   Note that the range of this sequence is finite, consisting of only the values 3 and 5. This sequence is plotted in Figure 9.1.4.(b).

3. \(\ds a_1= \frac{(-1)^{1(2)/2}}{1^2} = -1;\quad\) \(\ds a_2 = \frac{(-1)^{2(3)/2}}{2^2} =-\frac14 ;\quad\) \(\ds a_3 = \frac{(-1)^{3(4)/2}}{3^2} = \frac19;\quad\) \(\ds a_4 = \frac{(-1)^{4(5)/2}}{4^2} = \frac1{16};\quad\text{;}\) \(\ds a_5 = \frac{(-1)^{5(6)/2}}{5^2}=-\frac1{25}\quad\text{.}\)

   We gave one extra term to begin to show the pattern of signs is \(-\text{,}\) \(-\text{,}\) \(+\text{,}\) \(+\text{,}\) \(-\text{,}\) \(-\text{,}\) \(\ldots\text{,}\) due to the fact that the exponent of \(-1\) is a special quadratic. This sequence is plotted in Figure 9.1.4.(c).

   

   

   

We should first note that there is never exactly one function that describes a finite set of numbers as a sequence. There are many sequences that start with 2, then 5, as our first example does. We are looking for a simple formula that describes the terms given, knowing there is possibly more than one answer.

1. Note how each term is 3 more than the previous one. This implies a linear function would be appropriate: \(a(n) = a_n = 3n + b\) for some appropriate value of \(b\text{.}\) If we were to think in terms of ordered pairs, they would be of the form \((n,a(n))\text{.}\) So one such ordered pair would be \((1,2)\text{.}\) As we want \(a_1=2\text{,}\) we set \(b=-1\text{.}\) Thus \(a_n = 3n-1\text{.}\)

2. First notice how the sign changes from term to term. This is most commonly accomplished by multiplying the terms by either \((-1)^n\) or \((-1)^{n+1}\text{.}\) Using \((-1)^n\) multiplies the odd indexed terms by \((-1)\text{.}\) Thus the first term would be negative and the second term would be positive. Multiplying by \((-1)^{n+1}\) multiplies the even indexed terms by \((-1)\text{.}\) Thus the second term would be negative and the first term would be positive. As this sequence has negative even indexed terms, we will multiply by \((-1)^{n+1}\text{.}\)

   After this, we might feel a bit stuck as to how to proceed. At this point, we are just looking for a pattern of some sort among the absolute values of the terms: what do the numbers 2, 5, 10, 17, etc., have in common? There are many correct answers, but the one that we'll use here is that each is one more than a perfect square. That is, \(2=1^1+1\text{,}\) \(5=2^2+1\text{,}\) \(10=3^2+1\text{,}\) etc. Thus our formula is \(b_n= (-1)^{n+1}(n^2+1)\text{.}\)

3. One who is familiar with the factorial function will readily recognize these numbers. They are \(0!\text{,}\) \(1!\text{,}\) \(2!\text{,}\) \(3!\text{,}\) etc. Since our sequences start with \(n=1\text{,}\) we cannot write \(c_n = n!\text{,}\) for this misses the \(0!\) term. Instead, we shift by 1, and write \(c_n = (n-1)!\text{.}\)

4. This one may appear difficult, especially as the first two terms are the same, but a little “sleuthing” will help. Notice how the terms in the numerator are always multiples of 5, and the terms in the denominator are always powers of 2. Does something as simple as \(d_n = \frac{5n}{2^n}\) work?

   When \(n=1\text{,}\) we see that we indeed get \(5/2\) as desired. When \(n=2\text{,}\) we get \(10/4 = 5/2\text{.}\) Further checking shows that this formula indeed matches the other terms of the sequence.

1. Using Theorem 1.6.14, we can state that \(\lim\limits_{x\to\infty} \frac{3x^2-2x+1}{x^2-1000} = 3\text{.}\) (We could have also directly applied l'Hôpital's Rule.) Thus the sequence \(\{a_n\}\) converges, and its limit is 3. A scatter plot of every 5 values of \(a_n\) is given in Figure 9.1.10.(a). The values of \(a_n\) vary widely near \(n=30\text{,}\) ranging from about \(-73\) to \(125\text{,}\) but as \(n\) grows, the values approach 3.

2. The limit \(\lim\limits_{x\to\infty}\cos(x)\) does not exist, as \(\cos(x)\) oscillates (and takes on every value in \([-1,1]\) infinitely many times). Thus we cannot apply Theorem 9.1.8. The fact that the cosine function oscillates strongly hints that \(\cos(n)\text{,}\) when \(n\) is restricted to \(\mathbb{N}\text{,}\) will also oscillate. Figure 9.1.10.(b), where the sequence is plotted, shows that this is true. Because only discrete values of cosine are plotted, it does not bear strong resemblance to the familiar cosine wave. We conclude that \(\lim\limits_{n\to\infty}a_n\) does not exist.

3. We cannot actually apply Theorem 9.1.8 here, as the function \(f(x) = (-1)^x/x\) is not well defined. (What does \((-1)^{\sqrt{2}}\) mean? In actuality, there is an answer, but it involves complex analysis, beyond the scope of this text.) So for now we say that we cannot determine the limit. (But we will be able to very soon.) By looking at the plot in Figure 9.1.10.(c), we would like to conclude that the sequence converges to 0. That is true, but at this point we are unable to decisively say so.

1. This appeared in Example 9.1.9. We want to apply Theorem 9.1.11, so consider the limit of \(\{\abs{a_n}\}\text{:}\)

   \begin{align*} \lim_{n\to\infty} \abs{a_n} \amp = \lim_{n\to\infty} \abs{\frac{(-1)^n}{n}}\\ \amp = \lim_{n\to\infty} \frac{1}{n}\\ \amp = 0. \end{align*}

   Since this limit is 0, we can apply Theorem 9.1.11 and state that \(\lim\limits_{n\to\infty} a_n=0\text{.}\)

2. Because of the alternating nature of this sequence (i.e., every other term is multiplied by \(-1\)), we cannot simply look at the limit \(\lim\limits_{x\to\infty} \frac{(-1)^x(x+1)}{x}\text{.}\) We can try to apply the techniques of Theorem 9.1.11:

   \begin{align*} \lim_{n\to\infty} \abs{a_n} \amp = \lim_{n\to\infty} \abs{\frac{(-1)^n(n+1)}{n}}\\ \amp = \lim_{n\to\infty} \frac{n+1}{n}\\ \amp = 1. \end{align*}

   We have concluded that when we ignore the alternating sign, the sequence approaches 1. This means we cannot apply Theorem 9.1.11; it states the the limit must be 0 in order to conclude anything.

   

   Since we know that the signs of the terms alternate and we know that the limit of \(\abs{a_n}\) is 1, we know that as \(n\) approaches infinity, the terms will alternate between values close to 1 and \(-1\text{,}\) meaning the sequence diverges. A plot of this sequence is given in Figure 9.1.13.

We will use Theorem 9.1.14 to answer each of these.

1. Since \(\lim\limits_{n\to\infty} a_n = 0\) and \(\lim\limits_{n\to\infty} b_n = e\text{,}\) we conclude that \(\lim\limits_{n\to\infty} (a_n+b_n) = 0+e = e\text{.}\) So even though we are adding something to each term of the sequence \(b_n\text{,}\) we are adding something so small that the final limit is the same as before.

2. Since \(\lim\limits_{n\to\infty} b_n = e\) and \(\lim\limits_{n\to\infty} c_n = 5\text{,}\) we conclude that \(\lim\limits_{n\to\infty} (b_n\cdot c_n) = e\cdot 5 = 5e\text{.}\)

3. Since \(\lim\limits_{n\to\infty} a_n = 0\text{,}\) we have \(\lim\limits_{n\to\infty} 1000a_n =1000\cdot 0 = 0\text{.}\) It does not matter that we multiply each term by 1000; the sequence still approaches 0. (It just takes longer to get close to 0.)

1. The terms of this sequence are always positive but are decreasing, so we have \(0\lt a_n\lt 2\) for all \(n\text{.}\) Thus this sequence is bounded. Figure 9.1.18.(a)illustrates this.

2. The terms of this sequence obviously grow without bound. However, it is also true that these terms are all positive, meaning \(0\lt a_n\text{.}\) Thus we can say the sequence is unbounded, but also bounded below. Figure 9.1.18.(b) illustrates this.

In each of the following, we will examine \(a_{n+1}-a_n\text{.}\) If \(a_{n+1}-a_n \gt 0\text{,}\) we conclude that \(a_n\lt a_{n+1}\) and hence the sequence is increasing. If \(a_{n+1}-a_n\lt 0\text{,}\) we conclude that \(a_n\gt a_{n+1}\) and the sequence is decreasing. Of course, a sequence need not be monotonic and perhaps neither of the above will apply.

We also give a scatter plot of each sequence. These are useful as they suggest a pattern of monotonicity, but analytic work should be done to confirm a graphical trend.

1. \begin{align*} a_{n+1}-a_n \amp = \frac{n+2}{n+1} - \frac{n+1}{n}\\ \amp = \frac{(n+2)(n)-(n+1)^2}{(n+1)n}\\ \amp = \frac{-1}{n(n+1)} \\ \amp \lt 0 \text{ for all \(n\). } \end{align*}

   Since \(a_{n+1}-a_n\lt 0\) for all \(n\text{,}\) we conclude that the sequence is decreasing.

2. \begin{align*} a_{n+1}-a_n \amp = \frac{(n+1)^2+1}{n+2} - \frac{n^2+1}{n+1} \\ \amp = \frac{\big((n+1)^2+1\big)(n+1)- (n^2+1)(n+2)}{(n+1)(n+2)}\\ \amp = \frac{n^2+4n+1}{(n+1)(n+2)}\\ \amp \gt 0 \text{ for all \(n\). } \end{align*}

   Since \(a_{n+1}-a_n\gt 0\) for all \(n\text{,}\) we conclude the sequence is increasing.

3. We can clearly see in Figure 9.1.22 (c), where the sequence is plotted, that it is not monotonic. However, it does seem that after the first 4 terms it is decreasing. To understand why, perform the same analysis as done before: \(\ds \begin{aligned}a_{n+1}-a_n \amp = \frac{(n+1)^2-9}{(n+1)^2-10(n+1)+26} - \frac{n^2-9}{n^2-10n+26} \\ \amp = \frac{n^2+2n-8}{n^2-8n+17}-\frac{n^2-9}{n^2-10n+26}\\ \amp = \frac{(n^2+2n-8)(n^2-10n+26)-(n^2-9)(n^2-8n+17)}{(n^2-8n+17)(n^2-10n+26)}\\ \amp = \frac{-10n^2+60n-55}{(n^2-8n+17)(n^2-10n+26)}. \end{aligned}\) We want to know when this is greater than, or less than, 0. The denominator is always positive, therefore we are only concerned with the numerator. Using the quadratic formula, we can determine that \(-10n^2+60n-55=0\) when \(n\approx 1.13, 4.87\text{.}\) So for \(n\lt 1.13\text{,}\) the sequence is decreasing. Since we are only dealing with the natural numbers, this means that \(a_1 \gt a_2\text{.}\) Between \(1.13\) and \(4.87\text{,}\) i.e., for \(n=2\text{,}\) 3 and 4, we have that \(a_{n+1} \gt a_n\) and the sequence is increasing. (That is, when \(n=2\text{,}\) 3 and 4, the numerator \(-10n^2+60n+55\) from the fraction above is \(\gt 0\text{.}\)) When \(n\gt 4.87\text{,}\) i.e, for \(n\geq 5\text{,}\) we have that \(-10n^2+60n+55\lt 0\text{,}\) hence \(a_{n+1}-a_n\lt 0\text{,}\) so the sequence is decreasing. In short, the sequence is simply not monotonic. However, it is useful to note that for \(n\geq 5\text{,}\) the sequence is monotonically decreasing.

4. Again, the plot in Figure 9.1.(d) shows that the sequence is not monotonic, but it suggests that it is monotonically decreasing after the first term. We perform the usual analysis to confirm this.

   \begin{align*} a_{n+1}-a_n \amp = \frac{(n+1)^2}{(n+1)!} - \frac{n^2}{n!}\\ \amp = \frac{(n+1)^2-n^2(n+1)}{(n+1)!} \\ \amp = \frac{-n^3+2n+1}{(n+1)!} \end{align*}

   When \(n=1\text{,}\) the above expression is \(\gt 0\text{;}\) for \(n\geq 2\text{,}\) the above expression is \(\lt 0\text{.}\) Thus this sequence is not monotonic, but it is monotonically decreasing after the first term.