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Suppose that ${a}_{n}>0$ for all $n$ and that $\sum _{n=1}^{\infty}{a}_{n}$ converges. Suppose that ${b}_{n}$ is an arbitrary sequence of zeros and ones. Does $\sum _{n=1}^{\infty}{a}_{n}}{b}_{n$ necessarily converge?
Suppose that ${a}_{n}>0$ for all $n$ and that $\sum _{n=1}^{\infty}{a}_{n}$ diverges. Suppose that ${b}_{n}$ is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does $\sum _{n=1}^{\infty}{a}_{n}}{b}_{n$ necessarily diverge?
No. $\sum _{n=1}^{\infty}1\text{/}n$ diverges. Let ${b}_{k}=0$ unless $k={n}^{2}$ for some $n.$ Then $\sum _{k}{b}_{k}\text{/}k}={\displaystyle \sum 1\text{/}{k}^{2}$ converges.
Complete the details of the following argument: If $\sum _{n=1}^{\infty}\frac{1}{n}$ converges to a finite sum $s,$ then $\frac{1}{2}s=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\text{\cdots}$ and $s-\frac{1}{2}s=1+\frac{1}{3}+\frac{1}{5}+\text{\cdots}.$ Why does this lead to a contradiction?
Show that if ${a}_{n}\ge 0$ and $\sum _{n=1}^{\infty}{a}^{2}{}_{n}$ converges, then $\sum _{n=1}^{\infty}{\text{sin}}^{2}\left({a}_{n}\right)$ converges.
$\left|\text{sin}\phantom{\rule{0.1em}{0ex}}t\right|\le \left|t\right|,$ so the result follows from the comparison test.
Suppose that ${a}_{n}\text{/}{b}_{n}\to 0$ in the comparison test, where ${a}_{n}\ge 0$ and ${b}_{n}\ge 0.$ Prove that if $\sum {b}_{n}$ converges, then $\sum {a}_{n}$ converges.
Let ${b}_{n}$ be an infinite sequence of zeros and ones. What is the largest possible value of $x={\displaystyle \sum _{n=1}^{\infty}{b}_{n}\text{/}{2}^{n}}\text{?}$
By the comparison test, $x={\displaystyle \sum _{n=1}^{\infty}{b}_{n}\text{/}{2}^{n}}\le {\displaystyle \sum _{n=1}^{\infty}1\text{/}{2}^{n}}=1.$
Let ${d}_{n}$ be an infinite sequence of digits, meaning ${d}_{n}$ takes values in $\{0,\phantom{\rule{0.2em}{0ex}}1\text{,\u2026},9\}.$ What is the largest possible value of $x={\displaystyle \sum _{n=1}^{\infty}{d}_{n}\text{/}{10}^{n}}$ that converges?
Explain why, if $x>1\text{/}2,$ then $x$ cannot be written $x={\displaystyle \sum _{n=2}^{\infty}\frac{{b}_{n}}{{2}^{n}}}\left({b}_{n}=0\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}1,\phantom{\rule{0.2em}{0ex}}{b}_{1}=0\right).$
If ${b}_{1}=0,$ then, by comparison, $x\le {\displaystyle \sum _{n=2}^{\infty}1\text{/}{2}^{n}}=1\text{/}2.$
[T] Evelyn has a perfect balancing scale, an unlimited number of $1\phantom{\rule{0.2em}{0ex}}\text{-kg}$ weights, and one each of $1\text{/}2\phantom{\rule{0.2em}{0ex}}\text{-kg},\phantom{\rule{0.2em}{0ex}}1\text{/}4\phantom{\rule{0.2em}{0ex}}\text{-kg},\phantom{\rule{0.2em}{0ex}}1\text{/}8\phantom{\rule{0.2em}{0ex}}\text{-kg},$ and so on weights. She wishes to weigh a meteorite of unspecified origin to arbitrary precision. Assuming the scale is big enough, can she do it? What does this have to do with infinite series?
[T] Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of $1\phantom{\rule{0.2em}{0ex}}\text{-kg}$ weights, and nine each of $0.1\phantom{\rule{0.2em}{0ex}}\text{-kg,}$ $0.01\phantom{\rule{0.2em}{0ex}}\text{-kg},\phantom{\rule{0.2em}{0ex}}0.001\phantom{\rule{0.2em}{0ex}}\text{-kg,}$ and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?
Yes. Keep adding $1\phantom{\rule{0.2em}{0ex}}\text{-kg}$ weights until the balance tips to the side with the weights. If it balances perfectly, with Robert standing on the other side, stop. Otherwise, remove one of the $1\phantom{\rule{0.2em}{0ex}}\text{-kg}$ weights, and add $0.1\phantom{\rule{0.2em}{0ex}}\text{-kg}$ weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last $0.1\phantom{\rule{0.2em}{0ex}}\text{-kg}$ weight. Start adding $0.01\phantom{\rule{0.2em}{0ex}}\text{-kg}$ weights. If it balances, stop. If it tips to the side with the weights, remove the last $0.01\phantom{\rule{0.2em}{0ex}}\text{-kg}$ weight that was added. Continue in this way for the $0.001\phantom{\rule{0.2em}{0ex}}\text{-kg}$ weights, and so on. After a finite number of steps, one has a finite series of the form $A+{\displaystyle \sum _{n=1}^{N}{s}_{n}}\text{/}{10}^{n}$ where $A$ is the number of full kg weights and ${d}_{n}$ is the number of $1\text{/}{10}^{n}\phantom{\rule{0.2em}{0ex}}\text{-kg}$ weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the $N\text{th}$ partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most $1\text{/}{10}^{N}.$
The series $\sum _{n=1}^{\infty}\frac{1}{2n}$ is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which $n$ is odd. Let $m>1$ be fixed. Show, more generally, that deleting all terms $1\text{/}n$ where $n=mk$ for some integer $k$ also results in a divergent series.
In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A depleted harmonic series is a series obtained from $\sum _{n=1}^{\infty}\frac{1}{n}$ by removing any term $1\text{/}n$ if a given digit, say $9,$ appears in the decimal expansion of $n.$ Argue that this depleted harmonic series converges by answering the following questions.
a. ${10}^{d}-{10}^{d-1}<{10}^{d}$ b. $h\left(d\right)<{9}^{d}$ c. $m\left(d\right)={10}^{d-1}+1$ d. Group the terms in the deleted harmonic series together by number of digits. $h\left(d\right)$ bounds the number of terms, and each term is at most $1\text{/}m\left(d\right).$ $\sum _{d=1}^{\infty}h\left(d\right)\text{/}m\left(d\right)}\le {\displaystyle \sum _{d=1}^{\infty}{9}^{d}\text{/}{\left(10\right)}^{d-1}}\le 90.$ One can actually use comparison to estimate the value to smaller than $80.$ The actual value is smaller than $23.$
Suppose that a sequence of numbers ${a}_{n}>0$ has the property that ${a}_{1}=1$ and ${a}_{n+1}=\frac{1}{n+1}{S}_{n},$ where ${S}_{n}={a}_{1}+\text{\cdots}+{a}_{n}.$ Can you determine whether $\sum _{n=1}^{\infty}{a}_{n}$ converges? ( Hint: ${S}_{n}$ is monotone.)
Suppose that a sequence of numbers ${a}_{n}>0$ has the property that ${a}_{1}=1$ and ${a}_{n+1}=\frac{1}{{\left(n+1\right)}^{2}}{S}_{n},$ where ${S}_{n}={a}_{1}+\text{\cdots}+{a}_{n}.$ Can you determine whether $\sum _{n=1}^{\infty}{a}_{n}$ converges? ( Hint: ${S}_{2}={a}_{2}+{a}_{1}={a}_{2}+{S}_{1}={a}_{2}+1=1+1\text{/}4=\left(1+1\text{/}4\right){S}_{1},$ ${S}_{3}=\frac{1}{{3}^{2}}{S}_{2}+{S}_{2}=\left(1+1\text{/}9\right){S}_{2}=\left(1+1\text{/}9\right)\left(1+1\text{/}4\right){S}_{1},$ etc. Look at $\text{ln}\left({S}_{n}\right),$ and use $\text{ln}\left(1+t\right)\le t,$ $t>0.)$
Continuing the hint gives ${S}_{N}=\left(1+1\text{/}{N}^{2}\right)\left(1+1\text{/}{\left(N-1\right)}^{2}\text{\u2026}\left(1+1\text{/}4\right)\right).$ Then $\text{ln}\left({S}_{N}\right)=\text{ln}\left(1+1\text{/}{N}^{2}\right)+\text{ln}\left(1+1\text{/}{\left(N-1\right)}^{2}\right)+\text{\cdots}+\text{ln}\left(1+1\text{/}4\right).$ Since $\text{ln}\left(1+t\right)$ is bounded by a constant times $t,$ when $0<t<1$ one has $\text{ln}\left({S}_{N}\right)\le C{\displaystyle \sum _{n=1}^{N}\frac{1}{{n}^{2}}},$ which converges by comparison to the p -series for $p=2.$
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