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Question Number 131825 by liberty last updated on 09/Feb/21

\underset{n = 1}{\sum^{\infty}} \frac{2 n - 1}{2^{n}} ?

Answered by mr W last updated on 09/Feb/21

m e t h o d 1

\underset{n = 1}{\sum^{\infty}} x^{2 n} = \frac{x^{2}}{1 - x^{2}}

\underset{n = 1}{\sum^{\infty}} x^{2 n - 1} = \frac{x}{1 - x^{2}}

\underset{n = 1}{\sum^{\infty}} (2 n - 1) x^{2 n - 2} = \frac{1}{1 - x^{2}} + \frac{2 x^{2}}{{(1 - x^{2})}^{2}}

\underset{n = 1}{\sum^{\infty}} (2 n - 1) x^{2 n} = \frac{x^{2}}{1 - x^{2}} + \frac{2 x^{4}}{{(1 - x^{2})}^{2}}

x^{2} = \frac{1}{2}

\underset{n = 1}{\sum^{\infty}} \frac{2 n - 1}{2^{n}} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} + \frac{2 \times \frac{1}{4}}{{(1 - \frac{1}{2})}^{2}} = 1 + 2 = 3

Commented by liberty last updated on 09/Feb/21

cool \dots

Commented by mr W last updated on 09/Feb/21

m e t h o d 2 :

\underset{n = 1}{\sum^{\infty}} \frac{2 n - 1}{2^{n}}

= \underset{n = 1}{\sum^{\infty}} \frac{n}{2^{n - 1}} - \underset{n = 1}{\sum^{\infty}} \frac{1}{2^{n}} = S_{1} - S_{2}

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

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

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

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

\Rightarrow S_{1} = 2 \times 2 = 4

\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{2 n - 1}{2^{n}} = 4 - 1 = 3

Answered by Dwaipayan Shikari last updated on 09/Feb/21

M e t h o d ⌊ π ⌋ :)

S = x + 3 x^{2} + 5 x^{3} + 7 x^{4} + . .

S (1 - x) = x + 2 x^{2} + 2 x^{2} + \dots

S (1 - x) = \frac{2 x}{(1 - x)} - x \Rightarrow S = \frac{2 x}{{(1 - x)}^{2}} - \frac{x}{(1 - x)} (x = \frac{1}{2})

S = 4 - 1 = 3