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Question Number 1997 by prakash jain last updated on 29/Oct/15
A=1+(1/2)+(1/3)+.... B=1+(1/2)+(1/3)+.... A−B=?
$${A}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…. \\ $$$${B}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…. \\ $$$${A}−{B}=? \\ $$
Commented by Rasheed Soomro last updated on 30/Oct/15
If ′...′ represents ′non−ending′ A=1+(1/2)+(1/3)+.... ↕ ↕ ↕ ... B=1+(1/2)+(1/3)+.... A−B=(1−1)+((1/2)−(1/2))+((1/3)−(1/3))+...+((1/n)−(1/n))+... =0+0+0+... =0.∞ =Indeterminate
$${If}\:'…'\:\:\:{represents}\:'{non}−{ending}'\:\: \\ $$$${A}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…. \\ $$$$\:\:\:\:\:\:\:\:\updownarrow\:\:\:\:\:\updownarrow\:\:\:\:\:\:\:\updownarrow\:\:\:\:\:…\: \\ $$$${B}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…. \\ $$$${A}−{B}=\left(\mathrm{1}−\mathrm{1}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\right)+…+\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}}\right)+… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}+\mathrm{0}+\mathrm{0}+… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}.\infty \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:={Indeterminate} \\ $$
Answered by Yozzi last updated on 29/Oct/15
A and B both represent infinite harmonic series. Both series are then divergent and A−B is undefined.
$${A}\:{and}\:{B}\:{both}\:{represent}\:{infinite} \\ $$$${harmonic}\:{series}.\:{Both}\:{series}\:{are}\:{then} \\ $$$${divergent}\:{and}\:{A}−{B}\:{is}\:{undefined}.\: \\ $$
Commented by Filup last updated on 29/Oct/15
Can′t you say: A=B=1+(1/2)+(1/3)+... Let S=A=B ∴A−B=S−S=0?
$$\mathrm{Can}'\mathrm{t}\:\mathrm{you}\:\mathrm{say}: \\ $$$${A}={B}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+… \\ $$$$\mathrm{Let}\:\mathrm{S}={A}={B} \\ $$$$\therefore{A}−{B}={S}−{S}=\mathrm{0}? \\ $$
Commented by Filup last updated on 29/Oct/15
Similar to: S=1−1+1−1+... S_1 =(1−1)+(1−1)+...=0 S_2 =1−(1−1)−(1−1)−...=1 S_3 =0.5 ∵1−S=1−(1−1+1−1+...) ∴1−S=S 2S=1 S=0, 0.5, 1 Similar to: A=B=1+(1/2)+(1/3)+... (1) A−B=(1+(1/2)+(1/3)+...)−(1+(1/2)+(1/3)+...) A−B=(1−1)+((1/2)−(1/2))+((1/3)−(1/3))+... ∴A−B=0 or (2) S=A=B=1+(1/2)+(1/3)+... ∴A−B=S−S=0 ∴A−B=0
$$\mathrm{Similar}\:\mathrm{to}: \\ $$$${S}=\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+… \\ $$$${S}_{\mathrm{1}} =\left(\mathrm{1}−\mathrm{1}\right)+\left(\mathrm{1}−\mathrm{1}\right)+…=\mathrm{0} \\ $$$${S}_{\mathrm{2}} =\mathrm{1}−\left(\mathrm{1}−\mathrm{1}\right)−\left(\mathrm{1}−\mathrm{1}\right)−…=\mathrm{1} \\ $$$${S}_{\mathrm{3}} =\mathrm{0}.\mathrm{5} \\ $$$$\because\mathrm{1}−{S}=\mathrm{1}−\left(\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+…\right) \\ $$$$\therefore\mathrm{1}−{S}={S} \\ $$$$\mathrm{2}{S}=\mathrm{1} \\ $$$$ \\ $$$${S}=\mathrm{0},\:\mathrm{0}.\mathrm{5},\:\mathrm{1} \\ $$$$ \\ $$$${S}\mathrm{imilar}\:\mathrm{to}: \\ $$$${A}={B}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+… \\ $$$$\left(\mathrm{1}\right)\:{A}−{B}=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…\right)−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…\right) \\ $$$${A}−{B}=\left(\mathrm{1}−\mathrm{1}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\right)+… \\ $$$$\therefore{A}−{B}=\mathrm{0} \\ $$$$ \\ $$$${or} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:{S}={A}={B}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+… \\ $$$$\therefore{A}−{B}={S}−{S}=\mathrm{0} \\ $$$$ \\ $$$$\therefore{A}−{B}=\mathrm{0} \\ $$
Commented by Yozzi last updated on 29/Oct/15
How are we so certain that the series of each A and B have the same number of terms as n→∞? Each series is of the form Σ_(r=1) ^∞ (1/n). Thus, it is possible as you rightly said that A could equal B be at an arbitrarily large value of n. ∴ one possibility is ∃N∈N such that the partial sum Σ_(r=1) ^N (1/n) is both equal to A and B⇒A−B=0. However, what if the series each differ by at least one term so we have N≠M (N>M>3)where A=Σ_(r=1) ^N (1/r) and B=Σ_(r=1) ^M (1/r). In this case we may have for example A=1+(1/2)+(1/3)+(1/4)+(1/5)+...+(1/(1234567890))+(1/(1234567891))+(1/(1234567892)) A=Σ_(r=1) ^(1234567892) (1/r) and B=Σ_(r=1) ^(1234567891) ⇒A≠B⇒A−B≠0. I say A−B is undefined because n=∞ represents an arbitrarily large number of terms so that one doesn′t know for sure that A=B though A and B have identical notational representations.
$${How}\:{are}\:{we}\:{so}\:{certain}\:{that}\:{the}\:{series} \\ $$$${of}\:{each}\:{A}\:{and}\:{B}\:{have}\:{the}\:{same}\:{number} \\ $$$${of}\:{terms}\:{as}\:{n}\rightarrow\infty? \\ $$$${Each}\:{series}\:{is}\:{of}\:{the}\:{form} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}.\:{Thus},\:{it}\:{is}\:{possible}\:{as}\:{you}\:{rightly} \\ $$$${said}\:{that}\:{A}\:{could}\:{equal}\:{B}\:{be}\:{at}\:{an} \\ $$$${arbitrarily}\:{large}\:{value}\:{of}\:{n}.\: \\ $$$$\therefore\:{one}\:{possibility}\:{is}\:\exists{N}\in\mathbb{N}\:{such}\:{that} \\ $$$${the}\:{partial}\:{sum}\:\underset{{r}=\mathrm{1}} {\overset{{N}} {\sum}}\frac{\mathrm{1}}{{n}}\:{is}\:{both}\:{equal} \\ $$$${to}\:{A}\:{and}\:{B}\Rightarrow{A}−{B}=\mathrm{0}. \\ $$$${However},\:{what}\:{if}\:{the}\:{series}\:{each}\: \\ $$$${differ}\:{by}\:{at}\:{least}\:{one}\:{term}\:{so}\:{we}\:{have} \\ $$$${N}\neq{M}\:\left({N}>{M}>\mathrm{3}\right){where}\:{A}=\underset{{r}=\mathrm{1}} {\overset{{N}} {\sum}}\frac{\mathrm{1}}{{r}}\: \\ $$$${and}\:{B}=\underset{{r}=\mathrm{1}} {\overset{{M}} {\sum}}\frac{\mathrm{1}}{{r}}.\:{In}\:{this}\:{case}\:{we}\:{may}\:{have}\:{for}\:{example} \\ $$$${A}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+…+\frac{\mathrm{1}}{\mathrm{1234567890}}+\frac{\mathrm{1}}{\mathrm{1234567891}}+\frac{\mathrm{1}}{\mathrm{1234567892}} \\ $$$${A}=\underset{{r}=\mathrm{1}} {\overset{\mathrm{1234567892}} {\sum}}\frac{\mathrm{1}}{{r}} \\ $$$${and}\:{B}=\underset{{r}=\mathrm{1}} {\overset{\mathrm{1234567891}} {\sum}}\Rightarrow{A}\neq{B}\Rightarrow{A}−{B}\neq\mathrm{0}. \\ $$$${I}\:{say}\:{A}−{B}\:{is}\:{undefined}\:{because}\: \\ $$$${n}=\infty\:{represents}\:{an}\:{arbitrarily}\:{large}\: \\ $$$${number}\:{of}\:{terms}\:{so}\:{that}\:{one}\:{doesn}'{t} \\ $$$${know}\:{for}\:{sure}\:{that}\:{A}={B}\:{though} \\ $$$${A}\:{and}\:{B}\:{have}\:{identical}\:{notational} \\ $$$${representations}. \\ $$
Commented by Filup last updated on 29/Oct/15
I agree with the above. However, due to the similariy of a pattern and the open possible expectation that both patterns continue either the same or similarl magnitude, I would argue that both statements are correct such that: For: A=1+(1/2)+(1/3)+... A=Σ_(i=1) ^n (1/i)=H_n B=1+(1/2)+(1/3)+... B=Σ_(i=1) ^m (1/i)=H_m A−B= { ((0, n=m)),((undefined, n≠m)) :}
$$\mathrm{I}\:\mathrm{agree}\:\mathrm{with}\:\mathrm{the}\:\mathrm{above}.\:\mathrm{However},\:\mathrm{due}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{similariy}\:\mathrm{of}\:\mathrm{a}\:\mathrm{pattern}\:\mathrm{and}\:\mathrm{the}\:\mathrm{open} \\ $$$$\mathrm{possible}\:\mathrm{expectation}\:\mathrm{that}\:\mathrm{both}\:\mathrm{patterns}\:\mathrm{continue} \\ $$$$\mathrm{either}\:\mathrm{the}\:\mathrm{same}\:\mathrm{or}\:\mathrm{similarl}\:\mathrm{magnitude},\:\mathrm{I}\:\mathrm{would}\:\mathrm{argue} \\ $$$$\mathrm{that}\:\mathrm{both}\:\mathrm{statements}\:\mathrm{are}\:\mathrm{correct}\:\mathrm{such}\:\mathrm{that}: \\ $$$$ \\ $$$${For}: \\ $$$${A}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+… \\ $$$${A}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{i}}={H}_{{n}} \\ $$$$ \\ $$$${B}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+… \\ $$$${B}=\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\mathrm{1}}{{i}}={H}_{{m}} \\ $$$$ \\ $$$${A}−{B}=\begin{cases}{\mathrm{0},\:\:{n}={m}}\\{\mathrm{undefined},\:\:{n}\neq{m}}\end{cases} \\ $$$$ \\ $$
Commented by Yozzi last updated on 29/Oct/15
Correct.
$${Correct}. \\ $$
Commented by Rasheed Soomro last updated on 29/Oct/15
On the comment of Yozi You have objected about to be both series of same number of terms.I agree that one infinity may not be equal to other infinity however in this case I think both infinities are the same as of the infinity of N because they are concerned with counting. (Or all infinities are concerned with counting???) This can be said in better way that terms of any of the series can be matched with elements of N. Coudn′t we say that we are certain that both series have no end? when we consider that both series may have different number of terms we are really talking in finite sense_(−) .
$${On}\:{the}\:{comment}\:{of}\:{Yozi} \\ $$$${You}\:{have}\:{objected}\:{about}\:{to}\:{be}\:{both}\:{series}\: \\ $$$${of}\:{same}\:{number}\:{of}\:{terms}.{I}\:{agree}\:{that}\:{one}\:{infinity}\:{may}\:{not} \\ $$$${be}\:{equal}\:{to}\:{other}\:{infinity}\:{however}\:{in}\:{this}\:{case}\:{I}\:{think}\:{both} \\ $$$${infinities}\:{are}\:{the}\:{same}\:{as}\:{of}\:{the}\:{infinity}\:{of}\:\mathbb{N}\:{because} \\ $$$${they}\:{are}\:{concerned}\:{with}\:{counting}.\:\left({Or}\:{all}\:{infinities}\right. \\ $$$$\left.{are}\:{concerned}\:{with}\:{counting}???\right)\:{This}\:{can}\:{be}\:{said}\:{in}\:{better} \\ $$$${way}\:{that}\:{terms}\:{of}\:{any}\:{of}\:{the}\:{series}\:{can}\:{be}\:{matched}\:{with} \\ $$$${elements}\:{of}\:\mathbb{N}. \\ $$$${Coudn}'{t}\:{we}\:{say}\:{that}\:{we}\:{are}\:{certain}\:{that}\:{both}\:{series}\:{have} \\ $$$${no}\:{end}?\:{when}\:{we}\:{consider}\:{that}\:{both}\:{series}\:{may}\:{have}\: \\ $$$${different}\:{number}\:{of}\:{terms}\:{we}\:{are}\:{really}\:{talking}\:{in}\:\underset{−} {{finite}\:{sense}}. \\ $$
Commented by Yozzi last updated on 29/Oct/15
You′re correct, but then again if we say A=B can you prove to me that regardless of how large the terms of each of the series gets, A always equals B? The notation only suggests how each series is generated. The notation does not explicitly indicate whether or not the two series converge to each other. But, there′s a chance that I am misinterpretating(probably) how the idea of infinity plays out in deciding if A=B. My understanding of proving A=B is to show that the infinity concerned with A equals that concerned with B.
$${You}'{re}\:{correct},\:{but}\:{then}\:{again}\:{if}\:{we}\: \\ $$$${say}\:{A}={B}\:\:{can}\:{you}\:{prove}\:{to}\:{me}\:{that} \\ $$$${regardless}\:{of}\:{how}\:{large}\:{the}\:{terms} \\ $$$${of}\:{each}\:{of}\:{the}\:{series}\:{gets},\:{A}\:{always} \\ $$$${equals}\:{B}?\:{The}\:{notation}\:{only}\:{suggests} \\ $$$${how}\:{each}\:{series}\:{is}\:{generated}.\:{The} \\ $$$${notation}\:{does}\:{not}\:{explicitly}\:{indicate} \\ $$$${whether}\:{or}\:{not}\:{the}\:{two}\:{series}\:{converge} \\ $$$${to}\:{each}\:{other}.\:{But},\:{there}'{s}\:{a}\:{chance}\:{that} \\ $$$${I}\:{am}\:{misinterpretating}\left({probably}\right)\:{how}\:{the}\:{idea} \\ $$$${of}\:{infinity}\:{plays}\:{out}\:{in}\:{deciding}\:{if} \\ $$$${A}={B}.\:{My}\:{understanding}\:{of}\:{proving} \\ $$$${A}={B}\:{is}\:{to}\:{show}\:{that}\:{the}\:{infinity}\:{concerned} \\ $$$${with}\:{A}\:{equals}\:{that}\:{concerned}\:{with}\:{B}. \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 30/Oct/15
Pl see my comment following the Question. I have supported you even in case both infinities regardind A and B are same as of N.
$${Pl}\:{see}\:{my}\:{comment}\:{following}\:{the}\:{Question}. \\ $$$${I}\:{have}\:{supported}\:{you}\:{even}\:{in}\:{case}\:{both}\:{infinities} \\ $$$${regardind}\:{A}\:{and}\:{B}\:{are}\:{same}\:{as}\:{of}\:\mathbb{N}. \\ $$
Commented by 123456 last updated on 31/Oct/15
also if a series is conditionaly convergence you can change it value by reorganizing it a serie is conditionaly convergence if Σa_n converge but Σ∣a_n ∣ not if Σ∣a_n ∣ converge the covergence are called absolute covergenceu
$$\mathrm{also}\:\mathrm{if}\:\mathrm{a}\:\mathrm{series}\:\mathrm{is}\:\mathrm{conditionaly}\:\mathrm{convergence} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{change}\:\mathrm{it}\:\mathrm{value}\:\mathrm{by}\:\mathrm{reorganizing}\:\mathrm{it} \\ $$$$\mathrm{a}\:\mathrm{serie}\:\mathrm{is}\:\mathrm{conditionaly}\:\mathrm{convergence}\:\mathrm{if} \\ $$$$\Sigma{a}_{{n}} \:\mathrm{converge}\:\mathrm{but}\:\Sigma\mid{a}_{{n}} \mid\:\mathrm{not} \\ $$$$\mathrm{if}\:\Sigma\mid{a}_{{n}} \mid\:\mathrm{converge}\:\mathrm{the}\:\mathrm{covergence}\:\mathrm{are}\:\mathrm{called}\:\mathrm{absolute}\:\mathrm{covergenceu} \\ $$
Answered by Rasheed Soomro last updated on 01/Nov/15
A=1+(1/2)+(1/3)+.... ↕ ↕ ↕ .... B=1+(1/2)+(1/3)+.... A−B=(1−1)+((1/2)−(1/2))+...((1/n)−(1/n))+... =0+0+0+... =0.∞ =indeterminate
$$ \\ $$$${A}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…. \\ $$$$\:\:\:\:\:\:\:\:\updownarrow\:\:\:\:\:\updownarrow\:\:\:\:\:\:\:\updownarrow\:\:\:\:\:…. \\ $$$${B}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…. \\ $$$${A}−{B}=\left(\mathrm{1}−\mathrm{1}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+…\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}}\right)+… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}+\mathrm{0}+\mathrm{0}+… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}.\infty \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={indeterminate}\: \\ $$

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