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Compute \(1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.\)

Guest Jul 20, 2017
edited by Guest  Jul 20, 2017

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 #2
avatar+18369 
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Compute

\(1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.\)

 

Let \(r =\frac12\)

 

\(\begin{array}{|rcll|} \hline s_n &=& 1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} \\\\ && r = \frac12 \\\\ \hline s_n &=& 1\cdot r + 2\cdot r^2 + 3\cdot r^3 + \dots + n \cdot r^n \\ rs_n &=& \qquad \quad 1\cdot r^2 + 2\cdot r^3 + \dots + (n-1) \cdot r^n + n\cdot r^{n+1} \\ \hline s_n -r\cdot s_n &=& r+r^2+r^3+ \dots +r^n-n\cdot r^{n+1} \\ s_n\cdot(1-r) &=& \underbrace{( r+r^2+r^3+ \dots +r^n )}_{=S_n} -n\cdot r^{n+1} \\ \hline && S_n = r+r^2+r^3+ \dots +r^n \\ && rS_n = \quad r^2+r^3 + \dots + r^{n+1} \\ \hline && S_n -r\cdot S_n = r - r^{n+1} \\ && S_n\cdot (1-r) = r - r^{n+1} \\ && S_n = \frac{r - r^{n+1}}{1-r} \\ \hline s_n\cdot(1-r) &=& \underbrace{( r+r^2+r^3+ \dots +r^n )}_{=S_n} -n\cdot r^{n+1} \\ s_n\cdot(1-r) &=& \frac{r - r^{n+1}}{1-r} -n\cdot r^{n+1} \\ s_n &=& \frac{1}{1-r} \cdot \Big( \frac{r - r^{n+1}}{1-r} -n\cdot r^{n+1} \Big) \\\\ && \frac{1}{1-r} = 2 \\\\ s_n &=& 2 \cdot \Big( 2\cdot(r - r^{n+1}) -n\cdot r^{n+1} \Big) \\ s_n &=& 2 \cdot ( 2\cdot r - 2\cdot r^{n+1} - n\cdot r^{n+1} ) \\ s_n &=& 2 \cdot r\cdot ( 2 - 2\cdot r^n - n\cdot r^n ) \\ \\ \mathbf{s_n} & \mathbf{=} & \mathbf{2 \cdot r\cdot \Big( 2 - (2+n)\cdot r^n \Big)} \quad & | \quad r=\frac12 \\ \\ s_n & = & 2 \cdot \frac12 \cdot \Big( 2 - (2+n)\cdot (\frac12)^n \Big) \\ s_n & = & 2 - (2+n)\cdot \frac{1}{2^n} \\ \\ \mathbf{s_n} & \mathbf{=} & \mathbf{2 - \frac{2+n}{2^n} } \\ \hline \end{array}\)

 

laugh

heureka  Jul 21, 2017
edited by heureka  Jul 21, 2017
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2+0 Answers

 #1
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Your series can be summed up and it converges to 2 as follows:

 ∑[(1 / (2^n) * n), n, 1, 1000] =~2

Guest Jul 20, 2017
 #2
avatar+18369 
+1
Best Answer

Compute

\(1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.\)

 

Let \(r =\frac12\)

 

\(\begin{array}{|rcll|} \hline s_n &=& 1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} \\\\ && r = \frac12 \\\\ \hline s_n &=& 1\cdot r + 2\cdot r^2 + 3\cdot r^3 + \dots + n \cdot r^n \\ rs_n &=& \qquad \quad 1\cdot r^2 + 2\cdot r^3 + \dots + (n-1) \cdot r^n + n\cdot r^{n+1} \\ \hline s_n -r\cdot s_n &=& r+r^2+r^3+ \dots +r^n-n\cdot r^{n+1} \\ s_n\cdot(1-r) &=& \underbrace{( r+r^2+r^3+ \dots +r^n )}_{=S_n} -n\cdot r^{n+1} \\ \hline && S_n = r+r^2+r^3+ \dots +r^n \\ && rS_n = \quad r^2+r^3 + \dots + r^{n+1} \\ \hline && S_n -r\cdot S_n = r - r^{n+1} \\ && S_n\cdot (1-r) = r - r^{n+1} \\ && S_n = \frac{r - r^{n+1}}{1-r} \\ \hline s_n\cdot(1-r) &=& \underbrace{( r+r^2+r^3+ \dots +r^n )}_{=S_n} -n\cdot r^{n+1} \\ s_n\cdot(1-r) &=& \frac{r - r^{n+1}}{1-r} -n\cdot r^{n+1} \\ s_n &=& \frac{1}{1-r} \cdot \Big( \frac{r - r^{n+1}}{1-r} -n\cdot r^{n+1} \Big) \\\\ && \frac{1}{1-r} = 2 \\\\ s_n &=& 2 \cdot \Big( 2\cdot(r - r^{n+1}) -n\cdot r^{n+1} \Big) \\ s_n &=& 2 \cdot ( 2\cdot r - 2\cdot r^{n+1} - n\cdot r^{n+1} ) \\ s_n &=& 2 \cdot r\cdot ( 2 - 2\cdot r^n - n\cdot r^n ) \\ \\ \mathbf{s_n} & \mathbf{=} & \mathbf{2 \cdot r\cdot \Big( 2 - (2+n)\cdot r^n \Big)} \quad & | \quad r=\frac12 \\ \\ s_n & = & 2 \cdot \frac12 \cdot \Big( 2 - (2+n)\cdot (\frac12)^n \Big) \\ s_n & = & 2 - (2+n)\cdot \frac{1}{2^n} \\ \\ \mathbf{s_n} & \mathbf{=} & \mathbf{2 - \frac{2+n}{2^n} } \\ \hline \end{array}\)

 

laugh

heureka  Jul 21, 2017
edited by heureka  Jul 21, 2017

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