5.5 Alternating series  (Page 6/10)

Show that the alternating series $\frac{2}{3}-\frac{3}{5}+\frac{4}{7}-\frac{5}{9}+\text{⋯}$ does not converge. What hypothesis of the alternating series test is not met?

The formula $\text{cos}\theta =1-\frac{{\theta }^2}{2\text{!}}+\frac{{\theta }^4}{4\text{!}}-\frac{{\theta }^6}{6\text{!}}+\text{⋯}$ will be derived in the next chapter. Use the remainder $\left|R_N\right|\le b_{N+1}$ to find a bound for the error in estimating $\text{cos}\theta$ by the fifth partial sum $1-{\theta }^2\text{/}2\text{!}+{\theta }^4\text{/}4\text{!}\text{−}{\theta }^6\text{/}6\text{!}+{\theta }^8\text{/}8\text{!}$ for $\theta =1,$ $\theta =\pi \text{/}6,$ and $\theta =\pi .$

The formula $\text{sin}\theta =\theta -\frac{{\theta }^3}{3\text{!}}+\frac{{\theta }^5}{5\text{!}}-\frac{{\theta }^7}{7\text{!}}+\text{⋯}$ will be derived in the next chapter. Use the remainder $\left|R_N\right|\le b_{N+1}$ to find a bound for the error in estimating $\text{sin}\theta$ by the fifth partial sum $\theta -{\theta }^3\text{/}3\text{!}+{\theta }^5\text{/}5\text{!}\text{−}{\theta }^7\text{/}7\text{!}+{\theta }^9\text{/}9\text{!}$ for $\theta =1,$ $\theta =\pi \text{/}6,$ and $\theta =\pi .$

How many terms in $\text{cos}\theta =1-\frac{{\theta }^2}{2\text{!}}+\frac{{\theta }^4}{4\text{!}}-\frac{{\theta }^6}{6\text{!}}+\text{⋯}$ are needed to approximate $\text{cos}1$ accurate to an error of at most $0.00001\text{?}$

How many terms in $\text{sin}\theta =\theta -\frac{{\theta }^3}{3\text{!}}+\frac{{\theta }^5}{5\text{!}}-\frac{{\theta }^7}{7\text{!}}+\text{⋯}$ are needed to approximate $\text{sin}1$ accurate to an error of at most $0.00001\text{?}$

Sometimes the alternating series ${\displaystyle \sum _{n=1}^{\infty }{(\mathrm{-1})}^{n-1}}{b}_n$ converges to a certain fraction of an absolutely convergent series ${\displaystyle \sum _{n=1}^{\infty }{b}_n}$ at a faster rate. Given that ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}=\frac{{\pi }^2}{6},$ find $S=1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\text{⋯}.$ Which of the series $6{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}$ and $S{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(\mathrm{-1}\right)}^{n-1}}{n^2}}$ gives a better estimation of ${\pi }^2$ using $1000$ terms?

[T] $\frac{\pi }{4}={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(\mathrm{-1}\right)}^n}{2n+1}},$ error $<0.0001$

[T] $\frac{\pi }{\sqrt{12}}={\displaystyle \sum _{k=0}^{\infty }\frac{{\left(\mathrm{-3}\right)}^{\text{−}k}}{2k+1}},$ error $<0.0001$

[T] The series ${\displaystyle \sum _{n=0}^{\infty }\frac{\text{sin}\left(x+\pi n\right)}{x+\pi n}}$ plays an important role in signal processing. Show that ${\displaystyle \sum _{n=0}^{\infty }\frac{\text{sin}\left(x+\pi n\right)}{x+\pi n}}$ converges whenever $0<x<\pi .$ ( Hint: Use the formula for the sine of a sum of angles.)

[T] If ${\displaystyle \sum _{n=1}^N{\left(\mathrm{-1}\right)}^{n-1}\frac{1}{n}}\to \text{ln}2,$ what is $1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\frac{1}{8}-\frac{1}{10}-\frac{1}{12}+\text{⋯}\text{?}$

[T] Plot the series ${\displaystyle \sum _{n=1}^{100}\frac{\text{cos}\left(2\pi nx\right)}{n}}$ for $0\le x<1.$ Explain why ${\displaystyle \sum _{n=1}^{100}\frac{\text{cos}\left(2\pi nx\right)}{n}}$ diverges when $x=0,1.$ How does the series behave for other $x\text{?}$

[T] Plot the series ${\displaystyle \sum _{n=1}^{100}\frac{\text{sin}\left(2\pi nx\right)}{n}}$ for $0\le x<1$ and comment on its behavior

[T] Plot the series ${\displaystyle \sum _{n=1}^{100}\frac{\text{cos}(2\pi nx)}{n^2}}$ for $0\le x<1$ and describe its graph.

[T] The alternating harmonic series converges because of cancellation among its terms. Its sum is known because the cancellation can be described explicitly. A random harmonic series is one of the form ${\displaystyle \sum _{n=1}^{\infty }\frac{S_n}{n}},$ where $s_n$ is a randomly generated sequence of $\pm 1\text{'s}$ in which the values $\pm 1$ are equally likely to occur. Use a random number generator to produce $1000$ random $\pm 1\text{s}$ and plot the partial sums $S_N={\displaystyle \sum _{n=1}^N\frac{s_n}{n}}$ of your random harmonic sequence for $N=1$ to $1000.$ Compare to a plot of the first $1000$ partial sums of the harmonic series.

[T] Estimates of ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}$ can be accelerated by writing its partial sums as ${\displaystyle \sum _{n=1}^N\frac{1}{n^2}}={\displaystyle \sum _{n=1}^N\frac{1}{n(n+1)}}+{\displaystyle \sum _{n=1}^N\frac{1}{n^2(n+1)}}$ and recalling that ${\displaystyle \sum _{n=1}^N\frac{1}{n(n+1)}}=1-\frac{1}{N+1}$ converges to one as $N\to \infty .$ Compare the estimate of ${\pi }^2\text{/}6$ using the sums ${\displaystyle \sum _{n=1}^{1000}\frac{1}{n^2}}$ with the estimate using $1+{\displaystyle \sum _{n=1}^{1000}\frac{1}{n^2(n+1)}}.$

[T] The Euler transform rewrites $S={\displaystyle \sum _{n=0}^{\infty }{(\mathrm{-1})}^n}{b}_n$ as $S={\displaystyle \sum _{n=0}^{\infty }{(\mathrm{-1})}^n}{2}^{\text{−}n-1}{\displaystyle \sum _{m=0}^n\left(\begin{array}{c}n\\ m\end{array}\right)}{b}_{n-m}.$ For the alternating harmonic series, it takes the form $\text{ln}(2)={\displaystyle \sum _{n=1}^{\infty }\frac{{(\mathrm{-1})}^{n-1}}{n}}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n{2}^n}}.$ Compute partial sums of ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n{2}^n}}$ until they approximate $\text{ln}(2)$ accurate to within $0.0001.$ How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate $\text{ln}(2).$

[T] In the text it was stated that a conditionally convergent series can be rearranged to converge to any number. Here is a slightly simpler, but similar, fact. If $a_n\ge 0$ is such that $a_n\to 0$ as $n\to \infty$ but ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ diverges, then, given any number $A$ there is a sequence $s_n$ of $\pm 1\text{'s}$ such that ${\displaystyle \sum _{n=1}^{\infty }{a}_n}{s}_n\to A.$ Show this for $A>0$ as follows.

1. Recursively define $s_n$ by $s_n=1$ if $S_{n-1}={\displaystyle \sum _{k=1}^{n-1}{a}_k}{s}_k<A$ and $s_n=\mathrm{-1}$ otherwise.
2. Explain why eventually $S_n\ge A,$ and for any $m$ larger than this $n,$ $A-a_m\le S_m\le A+a_m.$
3. Explain why this implies that $S_n\to A$ as $n\to \infty .$