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Question Number 218438 by Marzuk last updated on 10/Apr/25
An amazing thing i saw  S = 1 + 2 + 3 + 4 + 5 + 6...      = 1 + 2(2/2 + 3/2 + 4/2 + 5/2 +6/2....)      = 1 + 2(1 + 3/2 + 2 + 5/2 + 3...)      = 1 + 2(1+ 2 + 3 ... + 3/2 + 5/2...)      = 1 + 2S + 2Σ_(n= 1) ^∞ ((2n + 1)/2)  or,S − 2S = 1 + Σ_(n=1) ^∞ 2n + 1    ∴ −S = Σ_(n=0) ^∞ 2n + 1  Sum of all odd numbers!  I know the step S−2S = −S is not allowed
$${An}\:{amazing}\:{thing}\:{i}\:{saw} \\ $$$${S}\:=\:\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:\mathrm{5}\:+\:\mathrm{6}… \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{2}/\mathrm{2}\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{4}/\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}\:+\mathrm{6}/\mathrm{2}….\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}\:+\:\mathrm{3}…\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{1}+\:\mathrm{2}\:+\:\mathrm{3}\:…\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}…\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}{S}\:+\:\mathrm{2}\underset{{n}=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}\:+\:\mathrm{1}}{\mathrm{2}} \\ $$$${or},{S}\:−\:\mathrm{2}{S}\:=\:\mathrm{1}\:+\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}{n}\:+\:\mathrm{1} \\ $$$$ \\ $$$$\therefore\:−{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}{n}\:+\:\mathrm{1} \\ $$$${Sum}\:{of}\:{all}\:{odd}\:{numbers}! \\ $$$${I}\:{know}\:{the}\:{step}\:{S}−\mathrm{2}{S}\:=\:−{S}\:{is}\:{not}\:{allowed} \\ $$
Answered by MrGaster last updated on 10/Apr/25
S≜Σ_(n=1) ^∞ n  ζ(s)≜Σ_(n=1) ^( ∞) n^(−s) (Re(s)>1)  Γ(s)ζ(s)=∫_0 ^∞ (x^(s−1) /(e^x −1))dx  Analytic continuation to s=1:  ζ(−1)=−(1/(12))  ⇒S=ζ(−1)=−(1/(12))    determinant (((S=−(1/(12)))))
$${S}\triangleq\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n} \\ $$$$\zeta\left({s}\right)\triangleq\underset{{n}=\mathrm{1}} {\overset{\:\infty} {\sum}}{n}^{−{s}} \left(\mathrm{Re}\left({s}\right)>\mathrm{1}\right) \\ $$$$\Gamma\left({s}\right)\zeta\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{s}−\mathrm{1}} }{{e}^{{x}} −\mathrm{1}}{dx} \\ $$$$\mathrm{Analytic}\:\mathrm{continuation}\:\mathrm{to}\:{s}=\mathrm{1}: \\ $$$$\zeta\left(−\mathrm{1}\right)=−\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\Rightarrow{S}=\zeta\left(−\mathrm{1}\right)=−\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\:\begin{array}{|c|}{{S}=−\frac{\mathrm{1}}{\mathrm{12}}}\\\hline\end{array} \\ $$
Commented by Marzuk last updated on 10/Apr/25
I am sorry but i am not able to understand  what are you saying because i really  dont know too much about zeta functions  integrations etc.I really like higher   concepts of math but i cant learn.  As English is my 2nd language and  most importantly my teachers never  helped me to learn higher concepts  there opinion is if i do, i will be mentally  mad.If they helped me i must study  complex analysis and differential geometry  right now.The concepts and research i know  and do is totally result of my own try.
$${I}\:{am}\:{sorry}\:{but}\:{i}\:{am}\:{not}\:{able}\:{to}\:{understand} \\ $$$${what}\:{are}\:{you}\:{saying}\:{because}\:{i}\:{really} \\ $$$${dont}\:{know}\:{too}\:{much}\:{about}\:{zeta}\:{functions} \\ $$$${integrations}\:{etc}.{I}\:{really}\:{like}\:{higher}\: \\ $$$${concepts}\:{of}\:{math}\:{but}\:{i}\:{cant}\:{learn}. \\ $$$${As}\:{English}\:{is}\:{my}\:\mathrm{2}{nd}\:{language}\:{and} \\ $$$${most}\:{importantly}\:{my}\:{teachers}\:{never} \\ $$$${helped}\:{me}\:{to}\:{learn}\:{higher}\:{concepts} \\ $$$${there}\:{opinion}\:{is}\:{if}\:{i}\:{do},\:{i}\:{will}\:{be}\:{mentally} \\ $$$${mad}.{If}\:{they}\:{helped}\:{me}\:{i}\:{must}\:{study} \\ $$$${complex}\:{analysis}\:{and}\:{differential}\:{geometry} \\ $$$${right}\:{now}.{The}\:{concepts}\:{and}\:{research}\:{i}\:{know} \\ $$$${and}\:{do}\:{is}\:{totally}\:{result}\:{of}\:{my}\:{own}\:{try}. \\ $$

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