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1-1-1-1-1-1-1-S-n-S-n-1-1-1-1-1-1-1-2S-n-2-2-2-subtracting-S-n-1-1-1-1-1-1-1-1-1-1-S-n-1-2-S-n-1-2-I-have-found-this-while-experim

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Question Number 103219 by Dwaipayan Shikari last updated on 13/Jul/20
1+1+1+1+1+1+1+....=S_n S_n =1+1+1+1+1+1+1+.... 2S_n = 2 + 2 + 2+....... .......... subtracting −S_n =1−1+1−1+1−1+1−1+1−1+.... −S_n =(1/2) S_n =−(1/2) I have found this while experiment . I know the sum diverges but is it pretty cool? Kindly rectify me if there is any fault on this non rigorous process I have found some Ramanujan proof S_n =1+2+3+4+5+6+7+... 4S_n = 4+ 8 + 12+... −3S_n =1−2+3−4+5−6+7−8+...... −3S_n =(1/4) S_n =−(1/(12)) Ramanujan had done this on his notebook
$$\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+….={S}_{{n}} \\ $$$${S}_{{n}} =\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+…. \\ $$$$\mathrm{2}{S}_{{n}} =\:\:\:\:\mathrm{2}\:+\:\:\:\:\:\:\mathrm{2}\:\:\:+\:\:\:\:\:\mathrm{2}+……. \\ $$$$……….\:{subtracting} \\ $$$$−{S}_{{n}} =\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+…. \\ $$$$−{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:{S}_{{n}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${I}\:{have}\:{found}\:{this}\:{while}\:{experiment}\:.\:{I}\:{know}\:{the}\:{sum}\:{diverges} \\ $$$${but}\:{is}\:{it}\:{pretty}\:{cool}?\: \\ $$$${Kindly}\:{rectify}\:{me}\:{if}\:{there}\:{is}\:{any}\:{fault}\:{on}\:{this}\:{non}\:{rigorous} \\ $$$${process} \\ $$$${I}\:{have}\:{found}\:{some}\:{Ramanujan}\:{proof} \\ $$$${S}_{{n}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+… \\ $$$$\mathrm{4}{S}_{{n}} =\:\:\:\:\:\mathrm{4}+\:\:\:\mathrm{8}\:\:\:+\:\mathrm{12}+…\:\:\:\:\: \\ $$$$−\mathrm{3}{S}_{{n}} =\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+\mathrm{5}−\mathrm{6}+\mathrm{7}−\mathrm{8}+…… \\ $$$$−\mathrm{3}{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${S}_{{n}} =−\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$ \\ $$$${Ramanujan}\:{had}\:{done}\:{this}\:{on}\:{his}\:{notebook} \\ $$
Commented by JDamian last updated on 13/Jul/20
I always have been told ∞−∞ is undefined. In this case S_n =∞ ⇒ 2S_n =∞ ⇒ S_n −2S_n = ∞−∞
$${I}\:{always}\:{have}\:{been}\:{told}\:\:\:\infty−\infty\:\:{is} \\ $$$${undefined}. \\ $$$${In}\:{this}\:{case}\:{S}_{{n}} =\infty\:\Rightarrow\:\mathrm{2}{S}_{{n}} =\infty\:\Rightarrow \\ $$$${S}_{{n}} −\mathrm{2}{S}_{{n}} =\:\infty−\infty \\ $$
Commented by Dwaipayan Shikari last updated on 13/Jul/20
I know what do you want to say sir.But If we don′t take S_(n ) as infinite then what will happen?
$${I}\:{know}\:{what}\:{do}\:{you}\:{want}\:{to}\:{say}\:{sir}.{But}\:{If}\:{we}\:{don}'{t}\:{take}\:{S}_{{n}\:} {as} \\ $$$${infinite}\:{then}\:{what}\:{will}\:{happen}? \\ $$
Commented by PRITHWISH SEN 2 last updated on 13/Jul/20
First of all ∞ is not a number so ∞−∞=0 or ∞+∞=2∞ is useless. ∞ is a conception it is a symbol which represents the number which is greater than any large number what you can imagine ∵ it is not any arithmetic number that is why no arithematic operations (such as +,−,÷,×) can be done with ∞. Now as far series is concerned the addition or substraction of two series can be down only when these two series converges to some finite values. Now there are large as well as smaller ∞ like for the set of integer and for the set of real numbers the ∞ of the former set(natural number set ) is smaller then the ∞ of the later set (real number set ).I think this much is enough for your doubt. Thank you.
$$\mathrm{First}\:\mathrm{of}\:\mathrm{all}\:\infty\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{number}\:\mathrm{so}\:\infty−\infty=\mathrm{0} \\ $$$$\mathrm{or}\:\infty+\infty=\mathrm{2}\infty\:\mathrm{is}\:\mathrm{useless}.\:\infty\:\mathrm{is}\:\mathrm{a}\:\mathrm{conception}\:\mathrm{it}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{symbol}\:\mathrm{which}\:\mathrm{represents}\:\mathrm{the}\:\mathrm{number}\:\mathrm{which}\: \\ $$$$\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{any}\:\mathrm{large}\:\mathrm{number}\:\mathrm{what}\:\mathrm{you}\:\mathrm{can}\:\mathrm{imagine} \\ $$$$\because\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{any}\:\mathrm{arithmetic}\:\mathrm{number}\:\mathrm{that}\:\mathrm{is}\:\mathrm{why}\:\: \\ $$$$\mathrm{no}\:\mathrm{arithematic}\:\mathrm{operations}\:\left(\mathrm{such}\:\mathrm{as}\:+,−,\boldsymbol{\div},×\right) \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{with}\:\infty. \\ $$$$\mathrm{Now}\:\mathrm{as}\:\mathrm{far}\:\mathrm{series}\:\mathrm{is}\:\mathrm{concerned}\:\mathrm{the}\:\mathrm{addition}\:\mathrm{or} \\ $$$$\mathrm{substraction}\:\mathrm{of}\:\mathrm{two}\:\mathrm{series}\:\mathrm{can}\:\mathrm{be}\:\mathrm{down}\:\mathrm{only}\: \\ $$$$\mathrm{when}\:\mathrm{these}\:\mathrm{two}\:\mathrm{series}\:\mathrm{converges}\:\mathrm{to}\:\mathrm{some}\:\mathrm{finite} \\ $$$$\mathrm{values}. \\ $$$$\mathrm{Now}\:\mathrm{there}\:\mathrm{are}\:\mathrm{large}\:\mathrm{as}\:\mathrm{well}\:\mathrm{as}\:\mathrm{smaller}\:\infty \\ $$$$\mathrm{like}\:\mathrm{for}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{integer}\:\:\mathrm{and}\:\mathrm{for}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{real} \\ $$$$\mathrm{numbers}\:\mathrm{the}\:\infty\:\mathrm{of}\:\mathrm{the}\:\mathrm{former}\:\mathrm{set}\left(\mathrm{natural}\:\mathrm{number}\:\mathrm{set}\:\right)\:\mathrm{is}\:\mathrm{smaller}\:\mathrm{then}\:\mathrm{the} \\ $$$$\infty\:\mathrm{of}\:\mathrm{the}\:\mathrm{later}\:\mathrm{set}\:\left(\mathrm{real}\:\mathrm{number}\:\mathrm{set}\:\right).\mathrm{I}\:\mathrm{think}\:\mathrm{this} \\ $$$$\mathrm{much}\:\mathrm{is}\:\mathrm{enough}\:\mathrm{for}\:\mathrm{your}\:\mathrm{doubt}.\:\mathrm{Thank}\:\mathrm{you}. \\ $$
Answered by prakash jain last updated on 13/Jul/20
ζ(0)=−(1/2),ζ(−1)=−(1/(12)) ζ(s) i think you already know. I think you should read the links that I sent earlier on this topic to get some insight on why it works. There a quite a bit of maths involved and I really dont want to retype material available. I admire your curiosity and if you put some more effort you will feel more comfortable when looking at sum of divergent series such as 1+2+3+4+...=−(1/(12))
$$\zeta\left(\mathrm{0}\right)=−\frac{\mathrm{1}}{\mathrm{2}},\zeta\left(−\mathrm{1}\right)=−\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\zeta\left({s}\right)\:\mathrm{i}\:\mathrm{think}\:\mathrm{you}\:\mathrm{already}\:\mathrm{know}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{should}\:\mathrm{read}\:\mathrm{the}\:\mathrm{links}\:\mathrm{that} \\ $$$$\mathrm{I}\:\mathrm{sent}\:\mathrm{earlier}\:\mathrm{on}\:\mathrm{this}\:\mathrm{topic}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{some}\:\mathrm{insight}\:\mathrm{on}\:\mathrm{why}\:\mathrm{it}\:\mathrm{works}. \\ $$$$\mathrm{There}\:\mathrm{a}\:\mathrm{quite}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{of}\:\mathrm{maths}\:\mathrm{involved} \\ $$$$\mathrm{and}\:\mathrm{I}\:\mathrm{really}\:\mathrm{dont}\:\mathrm{want}\:\mathrm{to}\:\mathrm{retype} \\ $$$$\mathrm{material}\:\mathrm{available}. \\ $$$$\mathrm{I}\:\mathrm{admire}\:\mathrm{your}\:\mathrm{curiosity}\:\mathrm{and}\:\mathrm{if}\:\mathrm{you} \\ $$$$\mathrm{put}\:\mathrm{some}\:\mathrm{more}\:\mathrm{effort}\:\mathrm{you}\:\mathrm{will}\:\mathrm{feel} \\ $$$$\mathrm{more}\:\mathrm{comfortable}\:\mathrm{when}\:\mathrm{looking} \\ $$$$\mathrm{at}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{divergent}\:\mathrm{series}\:\mathrm{such} \\ $$$$\mathrm{as}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…=−\frac{\mathrm{1}}{\mathrm{12}} \\ $$
Commented by prakash jain last updated on 13/Jul/20
I will add a simple example of gemometric series f(x)=(1/(1−x))=1+x+x^2 +... ∣x∣<1 using analytical continuity so can sum 1+2+2^2 +2^3 +..=(1/(1−2))=−1 or you can write S=1+2(1+2+2^2 +..)=1+2S S=−1 I am not adding any mathematical proofs here they are given in the links that i shared before.
$$\mathrm{I}\:\mathrm{will}\:\mathrm{add}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{example}\:\mathrm{of} \\ $$$$\mathrm{gemometric}\:\mathrm{series} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\:\:\:\mid{x}\mid<\mathrm{1} \\ $$$$\mathrm{using}\:\mathrm{analytical}\:\mathrm{continuity}\:\mathrm{so} \\ $$$$\mathrm{can}\:\mathrm{sum} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} +..=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}}=−\mathrm{1} \\ $$$$\mathrm{or}\:\mathrm{you}\:\mathrm{can}\:\mathrm{write} \\ $$$$\mathrm{S}=\mathrm{1}+\mathrm{2}\left(\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +..\right)=\mathrm{1}+\mathrm{2}{S} \\ $$$${S}=−\mathrm{1} \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{adding}\:\mathrm{any}\:\mathrm{mathematical} \\ $$$$\mathrm{proofs}\:\mathrm{here}\:\mathrm{they}\:\mathrm{are}\:\mathrm{given}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{links}\:\mathrm{that}\:\mathrm{i}\:\mathrm{shared}\:\mathrm{before}. \\ $$
Commented by Dwaipayan Shikari last updated on 13/Jul/20
I am always curious about these type of series.
$${I}\:{am}\:{always}\:{curious}\:{about}\:{these}\:{type}\:{of}\:{series}.\: \\ $$
Commented by Dwaipayan Shikari last updated on 13/Jul/20
Sir why two types of results are here 1−2+3−4+5−6+...=(1/4) but logically it approches to −∞ or 1−2+3−4+....=−1−1−1−1−1−...=−(1+1+1+1+1+1+..) =−(−(1/2))=(1/2) i haven′t read the article fully But it gives a overview that it is a arithmatic mean of some results I will read the full article Thanking you sir for giving me this source.
$$\mathrm{Sir}\:\:\mathrm{why}\:\mathrm{two}\:\mathrm{types}\:\mathrm{of}\:\mathrm{results}\:\mathrm{are}\:\mathrm{here}\: \\ $$$$ \\ $$$$\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+\mathrm{5}−\mathrm{6}+…=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{but}\:\mathrm{logically}\:\mathrm{it}\:\mathrm{approches}\:\mathrm{to}\:−\infty\: \\ $$$$\mathrm{or} \\ $$$$\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+….=−\mathrm{1}−\mathrm{1}−\mathrm{1}−\mathrm{1}−\mathrm{1}−…=−\left(\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+..\right) \\ $$$$=−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{i}\:\mathrm{haven}'\mathrm{t}\:\mathrm{read}\:\mathrm{the}\:\mathrm{article}\:\mathrm{fully} \\ $$$$\mathrm{But}\:\mathrm{it}\:\mathrm{gives}\:\mathrm{a}\:\mathrm{overview}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{arithmatic}\:\mathrm{mean}\:\mathrm{of}\:\mathrm{some} \\ $$$$\mathrm{results} \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{read}\:\mathrm{the}\:\mathrm{full}\:\mathrm{article} \\ $$$$\mathrm{Thanking}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{giving}\:\mathrm{me}\:\mathrm{this}\:\mathrm{source}. \\ $$
Commented by Dwaipayan Shikari last updated on 13/Jul/20
This is link which sir gave to me I want to share with everyone https://en.m.wikipedia.org/wiki/Analytic_continuation

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