What-is-the-value-of-this-infinite-sum-1-2-1-3-1-2-2-1-3-2-1-2-3-1-3-3-

Question Number 182810 by Mastermind last updated on 14/Dec/22

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{this}\:\mathrm{infinite}\:\mathrm{sum} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)+\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\right)+… \\ $$$$ \\ $$$$ \\ $$

Answered by JDamian last updated on 14/Dec/22

((1/2^1 )+(1/2^2 )+(1/2^3 )+∙∙∙)−((1/3^1 )+(1/3^2 )+(1/3^3 )+∙∙∙)= ((1/2)/(1−(1/2)))−((1/3)/(1−(1/3)))=(1/(2−1))−(1/(3−1))=1−(1/2)=(1/2)

$$\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\centerdot\centerdot\centerdot\right)−\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\centerdot\centerdot\centerdot\right)= \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}−\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{2}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}−\mathrm{1}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by HeferH last updated on 14/Dec/22

(1/2) + (1/2^2 ) + (1/2^3 ) + (1/2^4 )... = A (1/2)(1 + (1/2^1 ) + (1/2^2 ) + (1/2^3 )...) = A 1 +( (1/2^1 ) + (1/2^2 ) + (1/2^3 )...) = 2A 1 + A = 2A ⇒A = 1 (1/3) + (1/3^2 ) + (1/3^3 )... = B (1/3)(1 + (1/3) + (1/3^2 )...) = B 1 + ((1/3) + (1/3^2 )...) = 3B 1 + B = 3B ⇒ B = (1/2) for the question: ((1/2) − (1/3)) + ((1/2^2 ) − (1/3^2 )) + ((1/2^3 ) − (1/3^3 ))+ ... = C (1/2) + (1/2^2 ) + (1/2^3 ) ... − ((1/3) + (1/3^2 ) + (1/3^3 ) ...) = C 1 − (1/2) = C C = (1/2)

$$\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\:\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }…\:=\:{A} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }…\right)\:=\:{A} \\ $$$$\:\mathrm{1}\:+\left(\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }…\right)\:=\:\mathrm{2}{A} \\ $$$$\:\mathrm{1}\:+\:{A}\:=\:\mathrm{2}{A}\:\:\:\:\:\Rightarrow{A}\:=\:\mathrm{1} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }…\:=\:{B} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }…\right)\:=\:{B} \\ $$$$\:\mathrm{1}\:+\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }…\right)\:=\:\mathrm{3}{B} \\ $$$$\:\mathrm{1}\:+\:{B}\:=\:\mathrm{3}{B}\:\:\Rightarrow\:{B}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:{for}\:{the}\:{question}: \\ $$$$\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{3}}\right)\:+\:\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)\:+\:\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\:−\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\right)+\:…\:=\:{C} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\:\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\:…\:−\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\:…\right)\:=\:{C} \\ $$$$\:\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:{C} \\ $$$$\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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