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Question Number 125394 by Mammadli last updated on 10/Dec/20

$$\underset{\mathit{n}\to \mathrm{\infty }}{\mathit{l}\mathit{i}\mathit{m}}(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+...+\frac{2\mathit{n}-1}{2^{\mathit{n}}})=?$$

Answered by Dwaipayan Shikari last updated on 10/Dec/20

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2n-1}{2^n}=\sum ^{\mathrm{\infty }}\frac{n}{2^{n-1}}-\sum ^{\mathrm{\infty }}\frac{1}{2^n}$$$$\sum ^{\mathrm{\infty }}\frac{n}{2^{n-1}}=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+..=S$$$$S-\frac{S}{2}=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..\Rightarrow \frac{S}{2}=\frac{1}{1-\frac{1}{2}}\Rightarrow S=4$$$$\sum ^{\mathrm{\infty }}\frac{1}{2^n}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$$$$\sum ^{\mathrm{\infty }}\frac{2n-1}{2^n}=4-1=3$$

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