Chapter 9

Prove that if \(\lim _{n \rightarrow \infty} a_{n}=0\) and \(\left\\{b_{n}\right\\}\) is bounded then \(\lim _{n \rightarrow \infty} a_{n} b_{n}=0\).

Find the sum of the series $$ \sum_{k=1}^{\infty} \frac{2^{k}}{\left(2^{k+1}-1\right)\left(2^{k}-1\right)} $$

Prove that if \(\left\\{a_{n}\right\\}\) converges and \(\left\\{b_{n}\right\\}\) diverges then \(\left\\{a_{n}+b_{n}\right\\}\) diverges.

If \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) both diverge, does it follow that \(\left\\{a_{n}+b_{n}\right\\}\) diverges?

A famous sequence \(\left\\{f_{n}\right\\}\), called the Fibonacci Sequence after Leonardo Fibonacci, who introduced it around A.D. 1200 , is defined by the recursion formula $$f_{1}=f_{2}=1, \quad f_{n+2}=f_{n+1}+f_{n}$$ (a) Find \(f_{3}\) through \(f_{10}\). (b) Let \(\phi=\frac{1}{2}(1+\sqrt{5}) \approx 1.618034\). The Greeks called this number the golden ratio, claiming that a rectangle whose dimensions were in this ratio was "perfect." It can be shown that $$\begin{aligned} f_{n} &=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right] \\\ &=\frac{1}{\sqrt{5}}\left[\phi^{n}-(-1)^{n} \phi^{-n}\right] \end{aligned}$$ Check that this gives the right result for \(n=1\) and \(n=2\). The general result can be proved by induction (it is a nice challenge). More in line with this section, use this explicit formula to prove that \(\lim _{n \rightarrow \infty} f_{n+1} / f_{n}=\phi\). (c) Using the limit just proved, show that \(\phi\) satisfies the equation \(x^{2}-x-1=0\). Then, in another interesting twist, use the Quadratic Formula to show that the two roots of this equation are \(\phi\) and \(-1 / \phi\), two numbers that occur in the explicit formula for \(f_{n}\).

If an object of rest mass \(m_{0}\) has velocity \(v\), then (according to the theory of relativity) its mass \(m\) is given by \(m=\) \(m_{0} / \sqrt{1-v^{2} / c^{2}}\), where \(c\) is the velocity of light. Explain how physicists get the approximation $$ m \approx m_{0}+\frac{m_{0}}{2}\left(\frac{v}{c}\right)^{2} $$

Use the fact that \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow 0^{+}} f\left(\frac{1}{x}\right)\) to find the limits. $$ \lim _{n \rightarrow \infty}\left(1+\frac{1}{2 n}\right)^{n} $$

The author of a biology text claimed that the smallest positive solution to \(x=1-e^{-(1+k) x}\) is approximately \(x=2 k\), provided \(k\) is very small. Show how she reached this conclusion and check it for \(k=0.01\).

Use the fact that \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow 0^{+}} f\left(\frac{1}{x}\right)\) to find the limits. $$ \lim _{n \rightarrow \infty}\left(1+\frac{1}{n^{2}}\right)^{n} $$

Use the fact that \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow 0^{+}} f\left(\frac{1}{x}\right)\) to find the limits. $$ \lim _{n \rightarrow \infty}\left(\frac{n-1}{n+1}\right)^{n} $$

Let \(f(x)\) be a function that possesses at least \(n\) derivatives at \(x=a\) and let \(P_{n}(x)\) be the Taylor polynomial of order \(n\) based at \(a\). Show that $$ \begin{aligned} P_{n}(a)=f(a), \quad P_{n}^{\prime}(a)=f^{\prime}(a), \quad P_{n}^{\prime \prime}(a)=f^{\prime \prime}(a), \\ & \ldots, \quad P_{n}^{(n)}(a)=f^{(n)}(a) \end{aligned} $$

Use the fact that \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow 0^{+}} f\left(\frac{1}{x}\right)\) to find the limits. $$ \lim _{n \rightarrow \infty}\left(\frac{2+n^{2}}{3+n^{2}}\right)^{n} $$

Use the fact that \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow 0^{+}} f\left(\frac{1}{x}\right)\) to find the limits. $$ \lim _{n \rightarrow \infty}\left(\frac{2+n^{2}}{3+n^{2}}\right)^{n^{2}} $$

Show that if \(x\) is in \([0, \pi / 2]\) the error in using $$ \sin x \approx x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9 !} $$ is less than \(5 \times 10^{-5}\) and therefore, that this formula is good enough to build a four-place sine table.

Use Maclaurin's Formula, rather than l'Hôpital's Rule, to find (a) \(\lim _{x \rightarrow 0} \frac{\sin x-x+x^{3} / 6}{x^{5}}\) (b) \(\lim _{x \rightarrow 0} \frac{\cos x-1+x^{2} / 2-x^{4} / 24}{x^{6}}\)

Let \(g(x)=p(x)+x^{n+1} f(x)\), where \(p(x)\) is a polynomial of degree at most \(n\) and \(f\) has derivatives through order \(n\) Show that \(p(x)\) is the Maclaurin polynomial of order \(n\) for \(g\).

Recall that the Second Derivative Test for Local Extrema (Section 4.3) does not apply when \(f^{\prime \prime}(c)=0\). Prove the following generalization, which may help determine a maximum or a minimum when \(f^{\prime \prime}(c)=0 .\) Suppose that $$ f^{\prime}(c)=f^{\prime \prime}(c)=f^{\prime \prime \prime}(c)=\cdots=f^{(n)}(c)=0 $$ where \(n\) is odd and \(f^{(n+1)}(x)\) is continuous near \(c\). 1\. If \(f^{(n+1)}(c)<0\), then \(f(c)\) is a local maximum value. 2\. If \(f^{(n+1)}(c)>0\), then \(f(c)\) is a local minimum value. Test this result on \(f(x)=x^{4}\).