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A series is given by \(1/7^1+2/7^2+3/7^3+4/7^4+5/7^5+... .\)

Determine whether the series converges or diverges.

 Feb 21, 2020
 #1
avatar+109450 
+1

Each successive term is smaller than the preceding one......the series wil converge  ( to  7/36)

 

cool cool cool

 Feb 21, 2020
 #2
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does r=1/7?

Guest Feb 21, 2020
 #3
avatar+24388 
+1

A series is given by
\(\dfrac{1}{7^1}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\dfrac{4}{7^4}+\dfrac{5}{7^5}+\cdots \).

 

This is a infinite arithmetico-geometric sequence.

In general: \({\color{red}a}{\color{blue}b }+(a+d){\color{blue}br }+{\color{red}(a+2d)}{\color{blue}br^2 }+{\color{red}(a+3d)}{\color{blue}br^3 }+\cdots + {\color{red}\Big(a+(n-1)d \Big)}{\color{blue}br^{n-1} } + \cdots \)

 

\(\begin{array}{|lrcll|} \hline \text{arithmetical sequence:}\quad 1+2+3+4+5+\ldots \\ {\color{red}1}+({\color{red}1}+1*{\color{orange}1})+({\color{red}1}+2*{\color{orange}1})+({\color{red}1}+3*{\color{orange}1})+\dotsb+\Big({\color{red}1}+(n-1)*{\color{orange}1}\Big)+\dotsb \\ \boxed{a={\color{red}1},\ d={\color{orange}1} \text{ is the common difference} } \\\\ \text{geometric series:}\quad \frac{1}{7^1}+\frac{1}{7^2}+ \frac{1}{7^3}+ \frac{1}{7^4}+ \frac{1}{7^5}+\cdots \\ {\color{blue}\frac{1}{7} } +{\color{blue}\frac{1}{7} }\left({\color{green}\frac{1}{7}}\right)^1 +{\color{blue}\frac{1}{7} }\left({\color{green}\frac{1}{7}}\right)^2 +{\color{blue}\frac{1}{7} }\left({\color{green}\frac{1}{7}}\right)^3 +\dotsb +{\color{blue}\frac{1}{7} }\left({\color{green}\frac{1}{7}}\right)^{n-1} +\dotsb\\ \boxed{ b={\color{blue}\frac{1}{7} },\ r={\color{green}\frac{1}{7}}\ \text{ is the common ratio} } \\ \hline \end{array} \)

 

Formula:
sum of a infinite arithmetico-geometric sequence

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\\\ && \boxed{a={\color{red}1},\ d={\color{orange}1} \text{ is the common difference} } \\ &&\boxed{ b={\color{blue}\frac{1}{7} },\ r={\color{green}\frac{1}{7}}\ \text{ is the common ratio} } \\\\ s &=& \left(\dfrac{{\color{blue}\frac{1}{7}} }{1-{ \color{green}\frac{1}{7}} }\right) \Bigg( {\color{red}1} + { \color{orange}1} \left( \dfrac{ {\color{green}\frac{1}{7} }}{1-{\color{green}\frac{1}{7}} } \right) \Bigg) \\\\ s &=& \frac{1}{7}*\frac{7}{6} \left( {\color{red}1} + \frac{1}{7}*\frac{7}{6} \right) \\\\ s &=& \frac{1}{6} \left( {\color{red}1} + \frac{1}{6} \right) \\\\ s &=& \frac{1}{6} *\frac{7}{6} \\\\ \mathbf{s} &=& \mathbf{\frac{7}{36}} \\ \hline \end{array}\)

 

\(\mathbf{\dfrac{1}{7^1}+\dfrac{2}{7^2}+\dfrac{3}{7^3}+\dfrac{4}{7^4}+\dfrac{5}{7^5}+\cdots} = \mathbf{\dfrac{7}{36}}\)

 

laugh

 Feb 21, 2020

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