5.5 Alternating series  (Page 5/10)

${\displaystyle \sum }_{n=1}^{\infty }\frac{\text{cos}\left(n\pi \right)}{n}$

${\displaystyle \sum }_{n=1}^{\infty }\frac{\text{cos}\left(n\pi \right)}{n^{1\text{/}n}}$

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{n}\text{sin}\left(\frac{n\pi }{2}\right)$

${\displaystyle \sum }_{n=1}^{\infty }\text{sin}\left(n\pi \text{/}2\right)\text{sin}\left(1\text{/}n\right)$

[T] $b_n=1\text{/}n,$ error $<{10}^{\mathrm{-5}}$

[T] $b_n=1\text{/}\text{ln}\left(n\right),$ $n\ge 2,$ error $<{10}^{\mathrm{-1}}$

[T] $b_n=1\text{/}\sqrt{n},$ error $<{10}^{\mathrm{-3}}$

[T] $b_n=1\text{/}{2}^n,$ error $<{10}^{\mathrm{-6}}$

[T] $b_n=\text{ln}\left(1+\frac{1}{n}\right),$ error $<{10}^{\mathrm{-3}}$

[T] $b_n=1\text{/}{n}^2,$ error $<{10}^{\mathrm{-6}}$

If $b_n\ge 0$ is decreasing and $\underset{n\to \infty }{\text{lim}}{b}_n=0,$ then ${\displaystyle \sum _{n=1}^{\infty }\left(b_{2n-1}-b_{2n}\right)}$ converges absolutely.

If $b_n\ge 0$ is decreasing, then ${\displaystyle \sum _{n=1}^{\infty }\left(b_{2n-1}-b_{2n}\right)}$ converges absolutely.

If $b_n\ge 0$ and $\underset{n\to \infty }{\text{lim}}{b}_n=0$ then ${\displaystyle \sum _{n=1}^{\infty }(\frac{1}{2}(}{b}_{3n-2}+b_{3n-1})\text{−}{b}_{3n})$ converges.

If $b_n\ge 0$ is decreasing and ${\displaystyle \sum _{n=1}^{\infty }(b_{3n-2}+b_{3n-1}-b_{3n})}$ converges then ${\displaystyle \sum _{n=1}^{\infty }{b}_{3n-2}}$ converges.

If $b_n\ge 0$ is decreasing and ${\displaystyle \sum _{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n-1}{b}_n}$ converges conditionally but not absolutely, then $b_n$ does not tend to zero.

Let $a_n^{+}=a_n$ if $a_n\ge 0$ and $a_n^{-}=\text{−}{a}_n$ if $a_n<0.$ (Also, $a_n^{+}=0\text{if}{a}_n<0$ and $a_n^{-}=0\text{if}{a}_n\ge 0.)$ If ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ converges conditionally but not absolutely, then neither ${\displaystyle \sum _{n=1}^{\infty }{a}_n^{+}}$ nor ${\displaystyle \sum _{n=1}^{\infty }{a}_n^{-}}$ converge.

Suppose that $a_n$ is a sequence of positive real numbers and that ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ converges.

Suppose that $b_n$ is an arbitrary sequence of ones and minus ones. Does ${\displaystyle \sum _{n=1}^{\infty }{a}_n}{b}_n$ necessarily converge?

Suppose that $a_n$ is a sequence such that ${\displaystyle \sum _{n=1}^{\infty }{a}_n}{b}_n$ converges for every possible sequence $b_n$ of zeros and ones. Does ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ converge absolutely?

${\displaystyle \sum _{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\frac{{\text{sin}}^{2}n}{n}}$

${\displaystyle \sum _{n=1}^{\infty }{\left(\mathrm{-1}\right)}^{n+1}\frac{{\text{cos}}^{2}n}{n}}$

$1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\text{⋯}$

$1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\text{⋯}$

Show that the alternating series $1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{8}+\text{⋯}$ does

not converge. What hypothesis of the alternating series test is not met?

Suppose that ${\displaystyle \sum a_n}$ converges absolutely. Show that the series consisting of the positive terms $a_n$ also converges.