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Question Number 145809 by puissant last updated on 08/Jul/21

prove 1+2+3+.....=−(1/(12))

$$\mathrm{prove}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+.....=−\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Answered by Olaf_Thorendsen last updated on 09/Jul/21

A = 1−1+1−1+1−... A = 1−(1−1+1−1+1−...) A = 1−A A = 1/2 B = 1−2+3−4+5−6+7+... B = 1−(2−3+4−5+6−7+...) B = 1−(1−2+3−4+5−6+...) −(1−1+1−1+1−1+...) B = 1−B−A B = 1/4 S = 1+2+3+4+5+6+... S−B = (1+2+3+4+5+6+...) −( 1−2+3−4+5−6+7+...) S−B = 2∗(2+4+6+8+...) S−B = 4∗(1+2+3+4+...) = 4S S = −B/3 S = −1/12 !

$$\mathrm{A}\:=\:\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−... \\ $$$$\mathrm{A}\:=\:\mathrm{1}−\left(\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−...\right) \\ $$$$\mathrm{A}\:=\:\mathrm{1}−\mathrm{A} \\ $$$$\mathrm{A}\:=\:\mathrm{1}/\mathrm{2} \\ $$$$ \\ $$$$\mathrm{B}\:=\:\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+\mathrm{5}−\mathrm{6}+\mathrm{7}+... \\ $$$$\mathrm{B}\:=\:\mathrm{1}−\left(\mathrm{2}−\mathrm{3}+\mathrm{4}−\mathrm{5}+\mathrm{6}−\mathrm{7}+...\right) \\ $$$$\mathrm{B}\:=\:\mathrm{1}−\left(\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+\mathrm{5}−\mathrm{6}+...\right) \\ $$$$−\left(\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+...\right) \\ $$$$\mathrm{B}\:=\:\mathrm{1}−\mathrm{B}−\mathrm{A} \\ $$$$\mathrm{B}\:=\:\mathrm{1}/\mathrm{4} \\ $$$$ \\ $$$$\mathrm{S}\:=\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+... \\ $$$$\mathrm{S}−\mathrm{B}\:=\:\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+...\right) \\ $$$$−\left(\:\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+\mathrm{5}−\mathrm{6}+\mathrm{7}+...\right) \\ $$$$\mathrm{S}−\mathrm{B}\:=\:\mathrm{2}\ast\left(\mathrm{2}+\mathrm{4}+\mathrm{6}+\mathrm{8}+...\right) \\ $$$$\mathrm{S}−\mathrm{B}\:=\:\mathrm{4}\ast\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+...\right)\:=\:\mathrm{4S} \\ $$$$\mathrm{S}\:=\:−\mathrm{B}/\mathrm{3} \\ $$$$\mathrm{S}\:=\:−\mathrm{1}/\mathrm{12}\:! \\ $$

Commented by puissant last updated on 09/Jul/21

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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