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A-1-1-2-1-3-B-1-1-2-1-3-A-B-




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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