Coordinated Calculus

Since the formula is explicit, we simply substitute in the appropriate value for \(n\text{.}\)

1. Notice that \(s_1=2\) was given and we'll need it to find \(s_2\text{.}\) For \(n=2\text{,}\)

   \begin{equation*} s_2=s_{2-1}+2=s_1+2=2+2=4. \end{equation*}

   Similarly, to get \(s_3\) we'll need to know \(s_2\text{.}\)

   \begin{align*} s_3\amp=s_2+2=4+2=6,\\ s_4\amp=s_3+2=6+2=8,\\ s_5\amp=s_4+2=8+2=10,\\ s_6\amp=s_5+2=12\text{.} \end{align*}

2. This formula says that each term is the previous term multiplied by -2. We have

   \begin{align*} s_1\amp =1, \\ s_2\amp=-2(1)=-2, \\ s_3\amp=-2(-2)=4,\\ s_4\amp= -2(4)=-8, \\ s_5\amp=-2(-8)=16, \\ s_6\amp=-2(16)=-32 \text{.} \end{align*}

3. Notice that this formula relies on two previous terms, not just one, which is why we had to give the first two terms ahead of time.

   \begin{align*} s_1\amp=1,\\ s_2\amp=3,\\ s_3\amp=s_2 + 2 s_1 = 3+2(1)=5,\\ s_4\amp=s_3+2s_2 = 5 + 2(3)=11,\\ s_5\amp=s_4+2s_3 = 11 + 2(5)=21,\\ s_6\amp=s_5+2s_4 = 21+2(11) = 43\text{.} \end{align*}

1. When \(n=1\text{,}\) \(s_n = 1\text{;}\) when \(n=2\text{,}\) \(s_n = 2\text{.}\)

2. Think about the value of \(\frac{1}{n}\text{.}\)

3. Note that the numerator of each term of the sequence is one more than the denominator of the term.

1. \(s_n = n\text{.}\) A plot of the first 20 points in the sequence is shown below.

   

   This sequence does not have a limit as \(n\) goes to infinity.

2. \(s_n = \frac{1}{n}\text{.}\) A plot of the first 20 points in the sequence is shown below.

   

   This sequence has a limit of 0 as \(n\) goes to infinity.

3. \(s_n = \frac{n+1}{n}\text{.}\) A plot of the first 20 points in the sequence is shown below.

   

   This sequence has a limit of 1 as \(n\) goes to infinity.

1. By observation we see that a formula for \(s_n\) is \(s_n = n\text{.}\) A plot of the first 20 points in the sequence is shown below.

   

   We recall that a function \(f\) diverges to infinity (and hence does not have a limit³Limits when they exist are inherently finite. This is true even though we sometimes use the notation \(\lim\limits_{x\to\infty}f(x)=\infty\) to mean that a function increases without bound. as \(x\to\infty\)) if we can make the values of \(f(x)\) as large as we want by choosing large enough values of \(x\text{.}\) Since we can make the values of \(n\) in our sequence as large as we want by choosing \(n\) to be arbitrarily large, we suspect that this sequence does not have a limit as \(n\) goes to infinity. The fact that \(\lim\limits_{x\to\infty}x=\infty\) supports this reasoning.

2. By observation we see that a formula for \(s_n\) is \(s_n = \frac{1}{n}\text{.}\) A plot of the first 20 points in the sequence is shown below.

   

   Since we can make the values of \(\frac{1}{n}\) in our sequence as close to 0 as we want by choosing \(n\) to be arbitrarily large, we suspect that this sequence has a limit of 0 as \(n\) goes to infinity. This is supported by the fact that \(\lim\limits_{x\to\infty}\frac1x=0\text{.}\)

3. Since the numerator is always 1 more than the denominator, a formula for \(s_n\) is \(s_n = \frac{n+1}{n}\text{.}\) A plot of the first 20 points in the sequence is shown below.

   

   Since we can make the values of \(\frac{n+1}{n}\) in our sequence as close to 1 as we want by choosing \(n\) to be arbitrarily large, we suspect that this sequence has a limit of 1 as \(n\) goes to infinity. This is supported by the fact that \(\lim\limits_{x\to\infty}\frac{x+1}x=1\text{.}\)

1. A function \(f\) has limit \(L\) as \(x \to \infty\) if we can make \(f(x)\) as close to \(L\) as we like by \(\ldots\text{.}\)

2. Think about making \(s_n\) as close as you want to \(L\) for large values of \(n\text{.}\)

3. Consider the behavior of the function \(f(x) = \frac{1+x}{2+x}\) as \(x \to \infty\text{.}\)

1. A continuous function \(f\) has a limit \(L\) as the independent variable \(x\) goes to infinity if we can make the values of \(f(x)\) as close to \(L\) as we want by choosing large enough values of \(x\text{.}\)

2. We expect that a sequence \(\{s_n\}\) will have a limit \(L\) as \(n\) goes to infinity if we can make the entries \(s_n\) in the sequence as close to \(L\) as we want by choosing \(n\) to be sufficiently large.

3. As \(n\) gets large, the constant terms become infinitesimally small compared to \(n\) and so \(\frac{1+n}{2+n}\) looks like \(\frac{n}{n}\) or 1 for large \(n\text{.}\) So the sequence \(\left\{ \frac{1+n}{2+n}\right\}\) has a limit of 1 at infinity.

   More rigorously, we can either write \(\frac{1+n}{2+n}=1-\frac1{2+n}\) and note that \(\lim\limits_{n\to\infty}\frac1{2+n}=0\text{,}\) or we can apply L'Hopital's rule to the function \(f(x)=\frac{1+x}{2+x}\) as \(x\to\infty\) to assert that \(\left\{\frac{1+n}{2+n}\right\}\) has limit \(1\) as \(n\to\infty\text{.}\)

1. Multiply the numerator and denominator each by \(\frac{1}{n}\text{.}\)

2. Compare the \(n\)th term to \(\frac{3^n}{2^n}\text{.}\)

3. Plot the sequence and think about what the graph suggests.

1. The sequence \(\left\{\frac{1+2n}{3n-2}\right\}\) converges to \(\frac{2}{3}\text{.}\)

2. The sequence \(\left\{\frac{5+3^n}{10+2^n}\right\}\) diverges to infinity.

3. \(\frac{10^n}{n!} \to 0\) as \(n \to \infty\text{.}\)

1. A plot of the first 20 terms of the sequence \(\left\{\frac{1+2n}{3n-2}\right\}\) is shown below.

   

   The plot suggests that the sequence has a limit between 0.5 and 1. Evaluating the limit algebraically, we find that

   \begin{equation*} \lim\limits_{n\to\infty}\frac{1+2n}{3n-2}=\lim\limits_{n\to\infty}\frac{\frac1n+2}{3-\frac2n}=\frac23\text{.} \end{equation*}

   So the sequence \(\left\{\frac{1+2n}{3n-2}\right\}\) converges to \(\frac{2}{3}\text{.}\)

2. A plot of the first 20 terms of the sequence \(\left\{\frac{5+3^n}{10+2^n}\right\}\) is shown below. Note the scale on the vertical axis.

   

   The plot implies that the sequence does not have a limit as \(n\) goes to infinity. Evaluating the limit algebraically, we factor \(3^n\) from the numerator and \(2^n\) from the denominator. Doing so yields

   \begin{equation*} \lim\limits_{n\to\infty}\frac{5+3^n}{10+2^n}=\lim\limits_{n\to\infty}\frac{3^n\left(\frac{5}{3^n}+1\right)}{2^n\left(\frac{10}{2^n}+1\right)}\text{.} \end{equation*}

   Now, since \(\lim\limits_{n\to\infty}\frac{\frac5{3^n}+1}{\frac{10}{2^n}+1}=1\text{,}\) it follows that

   \begin{equation*} \lim\limits_{n\to\infty}\frac{3^n\left(\frac{5}{3^n}+1\right)}{2^n\left(\frac{10}{2^n}+1\right)}=\lim\limits_{n\to\infty}\left(\frac32\right)^n\text{.} \end{equation*}

   Since \(\frac{3}{2} \gt 1\text{,}\) this limit is infinite and the sequence \(\left\{\frac{5+3^n}{10+2^n}\right\}\) diverges to infinity as the plot suggests.

3. A plot of the first 20 terms of the sequence \(\left\{\frac{10^n}{n!}\right\}\) is shown below. Note the scale on the vertical axis.

   

   Initially, it looks as though the terms increase without bound, but beginning at about \(n=10\) the factorial in the denominator dominates the numerator. Notice that

   \begin{equation*} \frac{10^n}{n!} = \frac{10 \times 10 \times 10 \times \cdots \times 10}{1 \times 2 \times 3 \times \cdots \times n}\text{.} \end{equation*}

   When \(n \gt 20\text{,}\) we have that \(\frac{10}{n} \lt \frac{1}{2}\) and thus

   \begin{align*} \frac{10^n}{n!} \amp = \left(\frac{10 \times 10 \times 10 \times \cdots \times 10}{1 \times 2 \times 3 \times \cdots\times 20}\right) \left(\frac{10 \times 10 \times 10 \times \cdots \times 10}{21 \times 22 \times 23 \times \cdots\times n}\right)\\ \amp = \left(\frac{10^{20}}{20!}\right) \left(\frac{10}{21}\right) \left(\frac{10}{22}\right) \cdots \left(\frac{10}{n}\right)\\ \amp \lt \left(\frac{10^{20}}{20!}\right) \left(\frac{1}{2}\right)^{n-20}\text{.} \end{align*}

   Since \(\frac{1}{2}\lt 1\text{,}\) the term \(\left(\frac{1}{2}\right)^{n-20}\) goes to 0 as \(n\) goes to infinity. The fact that \(\frac{10^{20}}{20!}\) is a constant means that \(\frac{10^n}{n!} \to 0\) as \(n \to \infty\text{.}\)

1. This sequence is bounded because \(0 \leq \left(\frac{1}{3}\right)^n \leq \frac{1}{3} \) for all \(n\geq 1\text{.}\) It's also decreasing, so it's monotone as well. It converges and has limit 0.

2. This sequence is bounded because \(-1 \leq (-1)^n \leq 1 \) for all \(n\text{.}\) However, since \(s_n\) oscillates between \(-1\) and \(1\text{,}\) it is not monotone. The sequence diverges because of the oscillation.

3. \(2^n +1\) is unbounded because it gets arbitrarily large as \(n\) gets large. It is monotone increasing. The sequence diverges because it's unbounded.

4. This recursive sequence is bounded between \(0\) and \(30\text{,}\) and since each term is half the previous one, it is monotone decreasing. The sequence converges to \(0\text{.}\)

5. \((-2)^n \) is neither bounded nor monotone. It is not bounded below or above, and is not monotone since it alternates between positive and negative terms. The sequence diverges.

6. This sequence is not monotone, since it alternates between positive and negative terms. It is bounded between \(-1\) and \(1\) (also between \(-1\) and \(\frac12\)). This sequence converges to 0.

7. This sequence is both bounded and monotone. It's decreasing and bounded between 1 and 2. It converges with limit 1.