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

What is the value of this infinite sum

(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{2^{2}} - \frac{1}{3^{2}}) + (\frac{1}{2^{3}} - \frac{1}{3^{3}}) + \dots

Answered by JDamian last updated on 14/Dec/22

(\frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \cdot \cdot \cdot) - (\frac{1}{3^{1}} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \cdot \cdot \cdot) =

\frac{\frac{1}{2}}{1 - \frac{1}{2}} - \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2 - 1} - \frac{1}{3 - 1} = 1 - \frac{1}{2} = \frac{1}{2}

Answered by HeferH last updated on 14/Dec/22

\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \frac{1}{2^{4}} \dots = A

\frac{1}{2} (1 + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} \dots) = A

1 + (\frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} \dots) = 2 A

1 + A = 2 A \Rightarrow A = 1

\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} \dots = B

\frac{1}{3} (1 + \frac{1}{3} + \frac{1}{3^{2}} \dots) = B

1 + (\frac{1}{3} + \frac{1}{3^{2}} \dots) = 3 B

1 + B = 3 B \Rightarrow B = \frac{1}{2}

f o r t h e q u e s t i o n :

(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{2^{2}} - \frac{1}{3^{2}}) + (\frac{1}{2^{3}} - \frac{1}{3^{3}}) + \dots = C

\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} \dots - (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} \dots) = C

1 - \frac{1}{2} = C

C = \frac{1}{2}

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