Menu Close

show-that-1-2-3-4-1-8-

Tinku Tara 0 Comments
Pin




Question Number 176393 by BaliramKumar last updated on 18/Sep/22
show that 1+2+3+4+................ ∞ = ((−1)/8)
$${show}\:{that} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…………….\:\infty\:=\:\frac{−\mathrm{1}}{\mathrm{8}} \\ $$
Answered by BaliramKumar last updated on 19/Sep/22
S = 1+2+3+4+5+6+7+8+9+10+........∞ S = 1+2+3+4_(9) +5+6+7_(18) +8+9+10_(27) +........∞ S = 1+9+18+27+..............∞ S = 1 + 9(1+2+3+..........∞) S = 1 + 9(S) S − 9S = 1 −8S = 1 S = − (1/8)
$$\mathrm{S}\:=\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{10}+……..\infty \\ $$$$\mathrm{S}\:=\:\mathrm{1}+\underset{\mathrm{9}} {\underbrace{\mathrm{2}+\mathrm{3}+\mathrm{4}}}+\underset{\mathrm{18}} {\underbrace{\mathrm{5}+\mathrm{6}+\mathrm{7}}}+\underset{\mathrm{27}} {\underbrace{\mathrm{8}+\mathrm{9}+\mathrm{10}}}+……..\infty \\ $$$$\mathrm{S}\:=\:\mathrm{1}+\mathrm{9}+\mathrm{18}+\mathrm{27}+…………..\infty \\ $$$$\mathrm{S}\:=\:\mathrm{1}\:+\:\mathrm{9}\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+……….\infty\right) \\ $$$$\mathrm{S}\:=\:\mathrm{1}\:+\:\mathrm{9}\left(\mathrm{S}\right) \\ $$$$\mathrm{S}\:−\:\mathrm{9S}\:=\:\mathrm{1} \\ $$$$−\mathrm{8S}\:=\:\mathrm{1} \\ $$$$\mathrm{S}\:=\:−\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Commented by Mastermind last updated on 18/Sep/22
How did you got the last answer?
$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{got}\:\mathrm{the}\:\mathrm{last}\:\mathrm{answer}? \\ $$
Commented by Frix last updated on 19/Sep/22
funny. Ramanjuan showed S=−(1/(12)) and I saw a similar “proof” for S=+(1/(12)) S^∗ =1−1+1−1+... S^∗ =(1−1)+(1−1)+...=0+0+...=0 S^∗ =1+(−1+1)+(−1+1)+...=1+0+0+...=1 S^∗ =0 S^∗ =1 adding both 2S^∗ =1 S^∗ =(1/2) even more interesting is subtracting both which leads to 0=1 and the world implodes...
$$\mathrm{funny}. \\ $$$$\mathrm{Ramanjuan}\:\mathrm{showed}\:{S}=−\frac{\mathrm{1}}{\mathrm{12}}\:\mathrm{and}\:\mathrm{I}\:\mathrm{saw} \\ $$$$\mathrm{a}\:\mathrm{similar}\:“\mathrm{proof}''\:\mathrm{for}\:{S}=+\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${S}^{\ast} =\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+… \\ $$$${S}^{\ast} =\left(\mathrm{1}−\mathrm{1}\right)+\left(\mathrm{1}−\mathrm{1}\right)+…=\mathrm{0}+\mathrm{0}+…=\mathrm{0} \\ $$$${S}^{\ast} =\mathrm{1}+\left(−\mathrm{1}+\mathrm{1}\right)+\left(−\mathrm{1}+\mathrm{1}\right)+…=\mathrm{1}+\mathrm{0}+\mathrm{0}+…=\mathrm{1} \\ $$$${S}^{\ast} =\mathrm{0} \\ $$$${S}^{\ast} =\mathrm{1} \\ $$$$\mathrm{adding}\:\mathrm{both} \\ $$$$\mathrm{2}{S}^{\ast} =\mathrm{1} \\ $$$${S}^{\ast} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{even}\:\mathrm{more}\:\mathrm{interesting}\:\mathrm{is}\:\mathrm{subtracting}\:\mathrm{both} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{0}=\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{world}\:\mathrm{implodes}… \\ $$
Commented by Frix last updated on 19/Sep/22
S_1 =1+2+3+4+... S_2 =0+1+2+3+...^((∗)) S_1 −S_2 =1+1+1+1+... S_1 −S_2 =1+(1+1)+(1+1+1)+...=S_1 S_1 −S_2 =S_1 ⇒ S_2 =0 ⇒ S_1 +S_2 =S_1 S_1 +S_2 =1+3+5+7+...=−(1/8) 1+2+3+4+...=1+3+5+7+... ⇒ ⇒ S_3 =2+4+6+8+...=0 but S_3 =2S_1 ⇒ 0=−(1/4) ^((∗)) btw. S_2 =0+S_1 ⇔ 0=0−(1/8) ⇔ 0=−(1/8) ⇒ −(1/4)=−(1/8) ⇔ 4=8 ⇔ { ((0=4)),((1=2)) :} beautiful!
$${S}_{\mathrm{1}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+… \\ $$$${S}_{\mathrm{2}} =\mathrm{0}+\mathrm{1}+\mathrm{2}+\mathrm{3}+…\:^{\left(\ast\right)} \\ $$$${S}_{\mathrm{1}} −{S}_{\mathrm{2}} =\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+… \\ $$$${S}_{\mathrm{1}} −{S}_{\mathrm{2}} =\mathrm{1}+\left(\mathrm{1}+\mathrm{1}\right)+\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)+…={S}_{\mathrm{1}} \\ $$$${S}_{\mathrm{1}} −{S}_{\mathrm{2}} ={S}_{\mathrm{1}} \\ $$$$\Rightarrow\:{S}_{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{S}_{\mathrm{1}} +{S}_{\mathrm{2}} ={S}_{\mathrm{1}} \\ $$$${S}_{\mathrm{1}} +{S}_{\mathrm{2}} =\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+…=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…=\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+…\:\Rightarrow \\ $$$$\Rightarrow\:{S}_{\mathrm{3}} =\mathrm{2}+\mathrm{4}+\mathrm{6}+\mathrm{8}+…=\mathrm{0} \\ $$$$\mathrm{but} \\ $$$${S}_{\mathrm{3}} =\mathrm{2S}_{\mathrm{1}} \:\Rightarrow\:\mathrm{0}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:^{\left(\ast\right)} \:\mathrm{btw}.\:{S}_{\mathrm{2}} =\mathrm{0}+{S}_{\mathrm{1}} \:\Leftrightarrow\:\mathrm{0}=\mathrm{0}−\frac{\mathrm{1}}{\mathrm{8}}\:\Leftrightarrow\:\mathrm{0}=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\Rightarrow \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}=−\frac{\mathrm{1}}{\mathrm{8}}\:\Leftrightarrow\:\mathrm{4}=\mathrm{8}\:\Leftrightarrow\:\begin{cases}{\mathrm{0}=\mathrm{4}}\\{\mathrm{1}=\mathrm{2}}\end{cases} \\ $$$$\mathrm{beautiful}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *