APEX Calculus: for University of Lethbridge

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

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 10.1.5.

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

Note that the range of this sequence is finite, consisting of only the values 3 and 5. This sequence is plotted in Figure 10.1.6.

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

We gave one extra term to begin to show the pattern of signs is “\(-\text{,}\) \(-\text{,}\) \(+\text{,}\) \(+\text{,}\) \(-\text{,}\) \(-\text{,}\) \(\ldots\)”, due to the fact that the exponent of \(-1\) is a special quadratic. This sequence is plotted in Figure 10.1.7.

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 \(a_n = 3n-1\text{.}\)

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: 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, \(2=1^2+1\text{,}\) \(5=2^2+1\text{,}\) \(10=3^2+1\text{,}\) etc. Thus our formula is \(b_n= (-1)^{n+1}(n^2+1)\text{.}\)

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 \(c_n = (n-1)!\text{.}\)

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.

Using Theorem 1.6.21, 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’Hospital’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 10.1.14. 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.

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 10.1.12. 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 10.1.15, 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. The proof of the following statement is beyond the scope of this text, but it is true: there are infinitely many integers \(n\) that are arbitrarily (i.e., very) close to an even multiple of \(\pi\text{,}\) so that \(\cos n \approx 1\text{.}\) Similarly, there are infinitely many integers \(m\) that are arbitrarily close to an odd multiple of \(\pi\text{,}\) so that \(\cos m \approx -1\text{.}\) As the sequence takes on values near 1 and \(-1\) infinitely many times, we conclude that \(\lim\limits_{n\to\infty}a_n\) does not exist.

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 10.1.19.

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.

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

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.)

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 10.1.25.(a) illustrates this.

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 10.1.25.(b) illustrates this.

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. For small values of \(n\text{,}\) the numerator is positive. As \(n\) grows large, the numerator is dominated by \(-10n^2\text{,}\) meaning the entire fraction will be negative; i.e., for large enough \(n\text{,}\) \(a_{n+1}-a_n \lt 0\text{.}\) Using the quadratic formula we can determine that the numerator is negative for \(n\geq 5\text{.}\) In short, the sequence is simply not monotonic, though it is useful to note that for \(n\geq 5\text{,}\) the sequence is monotonically decreasing.