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Evaluate the sum \(\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{k}{2^k} + \cdots \)

 Jul 2, 2020
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Let

 \(\displaystyle S=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\frac{4}{2^{4}}+\dots +\frac{k}{2^{k}}+\dots \\=S_{1}+S_{2}+S_{3}+\dots+ S_{k}+\dots,\)

 

where

 

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

\(\displaystyle S_{2}=\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}+\dots.\)

\(\displaystyle S_{3}=\frac{1}{2^{3}}+\frac{1}{2^{4}}+\frac{1}{2^{5}}+\dots,\)

.

\(\displaystyle S_{k}=\frac{1}{2^{k}}+\frac{1}{2^{k+1}}+\frac{1}{2^{k+2}}+\dots,\)

.

.

Each   

\(\displaystyle S_{1},S_{2},S_{3}\dots,\)

is an infinite GP, (sum to infinity = a/(1 - r)) ) with common ratio 1/2. so,

\(\displaystyle S=\frac{1/2}{1-1/2}+\frac{1/2^{2}}{1-1/2}+\frac{1/2^{3}}{1-1/2}+\dots,\\ \displaystyle = 1+\frac{1}{2}+\frac{1}{4}+\dots,\)

which is again a GP common ratio 1/2.

\(\displaystyle S = 2.\)

 Jul 3, 2020

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