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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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