Chapter 11

For those sequences that converge, find the limit \(\lim _{n \rightarrow \infty} a_{n}\) a. \(a_{n}=\frac{n^{2}+1}{n^{3}+1} .\) b. \(a_{n}=\frac{3 n+1}{n+2}\). c. \(a_{n}=\left(\frac{3}{n}\right)^{1 / n}\). d. \(a_{n}=\frac{2 n^{2}+4 n^{3}}{n^{3}+5 \sqrt{2+n^{6}}} .\) e. \(a_{n}=n \ln \left(1+\frac{1}{n}\right)\). f. \(a_{n}=n \sin \left(\frac{1}{n}\right)\). g. \(a_{n}=\frac{(2 n+3) !}{(n+1) !} .\)

Find the sum for each of the series: a. \(\sum_{n=0}^{\infty} \frac{(-1)^{n} 3}{4^{n}}\) b. \(\sum_{n=2}^{\infty} \frac{2}{5^{n}}\). c. \(\sum_{n=0}^{\infty}\left(\frac{5}{2^{n}}+\frac{1}{3^{n}}\right)\). d. \(\sum_{n=1}^{\infty} \frac{3}{n(n+3)} .\)

Determine if the following converge, or diverge, using one of the convergence tests. If the series converges, is it absolute or conditional? a. \(\sum_{n=1}^{\infty} \frac{n+4}{2 n^{3}+1} .\) b. \(\sum_{n=1}^{\infty} \frac{\sin n}{n^{2}}\) c. \(\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^{2}}\). d. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{n-1}{2 n^{2}-3} .\) e. \(\sum_{n=1}^{\infty} \frac{\ln n}{n}\) f. \(\sum_{n=1}^{\infty} \frac{100^{n}}{n^{200}} .\) g. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{n}{n+3}\). h. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{\sqrt{5 n}}{n+1}\).

Do the following: a. Compute: \(\lim _{n \rightarrow \infty} n \ln \left(1-\frac{3}{n}\right)\). b. Use L'Hopital's Rule to evaluate \(L=\lim _{x \rightarrow \infty}\left(1-\frac{4}{x}\right)^{x}\). [Hint: Consider \(\ln L\).] c. Determine the convergence of \(\sum_{n=1}^{\infty}\left(\frac{n}{3 n+2}\right)^{n^{2}}\). d. Sum the series \(\sum_{n=1}^{\infty}\left[\tan ^{-1} n-\tan ^{-1}(n+1)\right]\) by first writing the \(N\) th partial sum and then computing \(\lim _{N \rightarrow \infty} s_{N}\).

Consider the sum \(\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}\) a. Use an appropriate convergence test to show that this series converges. b. Verify that $$ \sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}=\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right) $$ c. Find the \(n\)th partial sum of the series \(\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right)\) and use it to determine the sum of the resulting telescoping series.

Recall that the alternating harmonic series converges conditionally. a. From the Taylor series expansion for \(f(x)=\ln (1+x)\), inserting \(x=1\) gives the alternating harmonic series. What is the sum of the alternating harmonic series? b Because the alternating harmonic series does not converge absolutely, a rearrangement of the terms in the series will result in series whose sums vary. One such rearrangement in alternating \(p\) positive terms and \(n\) negative terms leads to the following sum \(^{10}\) : $$ \begin{gathered} \frac{1}{2} \ln \frac{4 p}{n}=\underbrace{\left(1+\frac{1}{3}+\cdots+\frac{1}{2 p-1}\right)}_{p \text { terms }}-\underbrace{\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right)}_{n \text { terms }} \\ +\underbrace{\left(\frac{1}{2 p+1}+\cdots+\frac{1}{4 p-1}\right)}_{p \text { terms }}-\underbrace{\left(\frac{1}{2 n+2}+\cdots+\frac{1}{4 n}\right)}_{n \text { terms }}+\cdots . \end{gathered} $$ Find rearrangements of the alternating harmonic series to give the following sums; that is, determine \(p\) and \(n\) for the given expression and write down the above series explicitly; that is, determine \(p\) and \(n\) leading to the following sums. i. \(\frac{5}{2} \ln 2\). ii. \(\ln 8\). iii. \(0 .\) iv. A sum that is close to \(\pi\).

Determine the radius and interval of convergence of the following infinite series: a. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{(x-1)^{n}}{n}\) b. \(\sum_{n=1}^{\infty} \frac{x^{n}}{2^{n} n !}\) c. \(\sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{x}{5}\right)^{n}\) d. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{x^{n}}{\sqrt{n}}\)

Find the Taylor series centered at \(x=a\) and its corresponding radius of convergence for the given function. In most cases, you need not employ the direct method of computation of the Taylor coefficients. a. \(f(x)=\sinh x, a=0\). b. \(f(x)=\sqrt{1+x}, a=0\). c. \(f(x)=x e^{x}, a=1\). d. \(f(x)=\frac{x-1}{2+x}, a=1\).

Test for pointwise and uniform convergence on the given set. [The Weierstraß M-Test might be helpful.] a. \(f(x)=\sum_{n=1}^{\infty} \frac{\ln n x}{n^{2}}, x \in[1,2]\) b. \(f(x)=\sum_{n=1}^{\infty} \frac{1}{3^{n}} \cos \frac{x}{2^{n}}\) on \(R\).

Consider Gregory's expansion $$ \tan ^{-1} x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\cdots=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2 k+1} x^{2 k+1} $$ a. Derive Gregory's expansion using the definition $$ \tan ^{-1} x=\int_{0}^{x} \frac{d t}{1+t^{2}} $$ expanding the integrand in a Maclaurin series, and integrating the resulting series term by term. b. From this result, derive Gregory's series for \(\pi\) by inserting an appropriate value for \(x\) in the series expansion for \(\tan ^{-1} x\).

Use deMoivre's Theorem to write \(\sin ^{3} \theta\) in terms of \(\sin \theta\) and \(\sin 3 \theta .\)

Evaluate the following expressions at the given point. Use your calculator and your computer (such as Maple). Then use series expansions to find an approximation to the value of the expression to as many places as you trust. a. \(\frac{1}{\sqrt{1+x^{3}}}-\cos x^{2}\) at \(x=0.015\). b. \(\ln \sqrt{\frac{1+x}{1-x}}-\tan x\) at \(x=0.0015\). c. \(f(x)=\frac{1}{\sqrt{1+2 x^{2}}}-1+x^{2}\) at \(x=5.00 \times 10^{-3}\). d. \(f(R, h)=R-\sqrt{R^{2}+h^{2}}\) for \(R=1.374 \times 10^{3} \mathrm{~km}\) and \(h=1.00 \mathrm{~m}\). e. \(f(x)=1-\frac{1}{\sqrt{1-x}}\) for \(x=2.5 \times 10^{-13}\)

Determine the order, \(O\left(x^{p}\right)\), of the following functions. You may need to use series expansions in powers of \(x\) when \(x \rightarrow 0\), or series expansions in powers of \(1 / x\) when \(x \rightarrow \infty\) a. \(\sqrt{x(1-x)}\) as \(x \rightarrow 0\). b. \(\frac{x^{5 / 4}}{1-\cos x}\) as \(x \rightarrow 0\) c. \(\frac{x}{x^{2}-1}\) as \(x \rightarrow \infty\). d. \(\sqrt{x^{2}+x}-x\) as \(x \rightarrow \infty\).