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Consider the series. 

 

\(\sum_{n+1}^{∞}\) (4n+1/4^(n+1))

 

What is the value of R?

Does the series converge or diverge?

 Feb 24, 2020
 #1
avatar+30673 
+5

The sum converges to 19/36.  Here's one possible derivation of it:

 

 Feb 24, 2020
 #2
avatar+25529 
+3

Consider the series. 

\(\sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n+1}}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n+1}}} &=& \sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n}*4} \\\\ &=& \dfrac{1}{4} \left( \sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n}} \right) \\\\ &=& \dfrac{1}{4} \left[ \sum \limits_{n=1}^{\infty} (1+4n) \left( \dfrac{1}{4}\right)^{n} \right] \\\\ &=& \dfrac{1}{4} \left[ -1 + \sum \limits_{n=1}^{\infty} \Big(1+4(n-1)\Big) \left( \dfrac{1}{4}\right)^{n-1} \right] \\\\ \mathbf{\sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n+1}}} &=& \mathbf{\dfrac{1}{4}( -1 + s )} \qquad \boxed{s=\sum \limits_{n=1}^{\infty} \Big(1+4(n-1)\Big) \left( \dfrac{1}{4}\right)^{n-1} } \\ \hline \end{array}\)

 

Infinite arithmetico-geometric sequence:

\(\begin{array}{|rcll|} \hline s &=& \sum \limits_{n=1}^{\infty} \Big(1+4(n-1)\Big) \left( \dfrac{1}{4}\right)^{n-1} \\ \hline \end{array}\)

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

 

\(\begin{array}{|lrcll|} \hline \text{arithmetical sequence:}\quad 1+5+9+13+17+\ldots \\ {\color{red}1}+({\color{red}1}+1*{\color{orange}4})+({\color{red}1}+2*{\color{orange}4})+({\color{red}1}+3*{\color{orange}4})+\dotsb+\Big({\color{red}1}+(n-1)*{\color{orange}4}\Big)+\dotsb \\ \boxed{a={\color{red}1},\ d={\color{orange}4} \text{ is the common difference} } \\\\ \text{geometric series:}\quad 1+1*\frac{1}{4^1}+1*\frac{1}{4^2}+ 1*\frac{1}{4^3}+ 1*\frac{1}{4^4}+\cdots \\ {\color{blue}1} +{\color{blue}1}*\left({\color{green}\frac{1}{4}}\right)^1 +{\color{blue}1}*\left({\color{green}\frac{1}{4}}\right)^2 +{\color{blue}1}*\left({\color{green}\frac{1}{4}}\right)^3 +\dotsb +{\color{blue}1}*\left({\color{green}\frac{1}{4}}\right)^{n-1} +\dotsb\\ \boxed{ b={\color{blue}1},\ r={\color{green}\frac{1}{4}}\ \text{ is the common ratio} } \\ \hline \end{array} \)

 

Formula:
sum of a infinite arithmetico-geometric sequence: \(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+d\left( \dfrac{r}{1-r} \right) \Bigg)} \\\\ && \boxed{a={\color{red}1},\ d={\color{orange}4} \text{ is the common difference} } \\ &&\boxed{ b={\color{blue}1 },\ r={\color{green}\frac{1}{4}}\ \text{ is the common ratio} } \\\\ s &=& \left(\dfrac{{\color{blue}1} }{1-{ \color{green}\frac{1}{4}} }\right) \Bigg( {\color{red}1} + { \color{orange}4} \left( \dfrac{ {\color{green}\frac{1}{4} }}{1-{\color{green}\frac{1}{4}} } \right) \Bigg) \\\\ s &=& \dfrac{4}{3}\left( {\color{red}1} + \dfrac{4}{3} \right) \\\\ s &=& \dfrac{4}{3} *\dfrac{7}{3} \\\\ \mathbf{s} &=& \mathbf{\dfrac{28}{9}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n+1}}} &=& \mathbf{\dfrac{1}{4}( -1 + s )} \quad | \quad s=\dfrac{28}{9} \\\\ &=& \dfrac{1}{4}\left( -1 + \dfrac{28}{9} \right) \\\\ &=& \dfrac{1}{4}\left( \dfrac{19}{9} \right) \\\\ \mathbf{\sum \limits_{n=1}^{\infty} \dfrac{1+4n} {4^{n+1}}} &=& \mathbf{\dfrac{19}{36}} \\ \hline \end{array}\)

 

laugh

 Feb 24, 2020

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