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Evaluate the infinite geometric series:\(1-\frac{2}{7}+\frac{4}{49}-\frac{8}{343}+\dots\)

 Apr 26, 2020
 #1
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0

sumfor(n, 1, 100,  (-1)^(n + 1)* (7/2)^(1 - n) = it converges to  7 / 9

 Apr 26, 2020
 #2
avatar+24995 
+3

Evaluate the infinite geometric series: \(1-\dfrac{2}{7}+\dfrac{4}{49}-\dfrac{8}{343}+\dots\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{1-\dfrac{2}{7}+\dfrac{4}{49}-\dfrac{8}{343}+\ldots} \\\\ &=& 1+\left(-\dfrac{2}{7}\right) +\left(-\dfrac{2}{7}\right)^2 +\left(-\dfrac{2}{7}\right)^3 +\ldots \quad | \quad \text{ratio}\ = -\dfrac{2}{7} \\ \hline \end{array}\)

 

Formula:

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\dfrac{1}{1-r}} \quad | \quad r=\text{ratio},\ s=\text{infinite sum} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\dfrac{1}{1-r}} \quad | \quad r = -\dfrac{2}{7} \\\\ s &=& \dfrac{1}{1-\left(-\dfrac{2}{7}\right)} \\\\ s &=& \dfrac{1}{1+ \dfrac{2}{7} } \\\\ s &=& \dfrac{1}{\dfrac{7+2}{7} } \\\\ s &=& \dfrac{7} {7+2} \\\\ \mathbf{s} &=& \mathbf{\dfrac{7}{9}} \\ \hline \end{array}\)

 

laugh

 Apr 27, 2020
 #3
avatar+633 
+1

Here is the proof for the formula.

AnExtremelyLongName  Apr 27, 2020

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