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Evaluate the sum \(1 + \frac{3}{3} + \frac{5}{9} + \frac{7}{27} + \frac{9}{81} + \dotsb\)

 Apr 29, 2019
 #1
avatar+217 
0

Because the sum goes on forever, we know that the sum from 5/9 on is 1, so 1 + 1 + 1 = 3

 Apr 29, 2019
 #2
avatar+24983 
+3

Evaluate the sum

\(\large{s= 1 + \dfrac{3}{3} + \dfrac{5}{9} + \dfrac{7}{27} + \dfrac{9}{81} + \dotsb }\)

 

\(\begin{array}{|rcll|} \hline s &=& 1 + \dfrac{3}{3} + \dfrac{5}{9} + \dfrac{7}{27} + \dfrac{9}{81} + \dotsb \\\\ &=& \dfrac{1}{3^0} + \dfrac{3}{3^1} + \dfrac{5}{3^2} + \dfrac{7}{3^3} + \dfrac{9}{3^4} + \dotsb + \dfrac{2n-1}{3^{n-1}} + \dotsb \\\\ \hline s_n &=& \dfrac{1}{3^0} + \dfrac{3}{3^1} + \dfrac{5}{3^2} + \dfrac{7}{3^3} + \dfrac{9}{3^4} + \dotsb + \dfrac{2n-1}{3^{n-1}} \\\\ &=& 1\cdot 3^0 + 3\cdot 3^{-1} + 5\cdot 3^{-2} + 7\cdot 3^{-3} + 9\cdot 3^{-4} + \dotsb + (2n-1)\cdot 3^{1-n} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline s_n &=& 1\cdot 3^0 + & 3\cdot 3^{-1} + 5\cdot 3^{-2} + 7\cdot 3^{-3} + 9\cdot 3^{-4} + \dotsb + (2n-1)\cdot 3^{-n+1} \\ 3^{-1} s_n &=& & 1\cdot 3^{-1} + 3\cdot 3^{-2} + 5\cdot 3^{-3} + 7\cdot 3^{-4} + \dotsb + (2n-3)\cdot 3^{-n+1}+ (2n-1)\cdot 3^{-n} \\ \hline \end{array} \\ \begin{array}{rcll} s_n-3^{-1} s_n &=& 1\cdot 3^0 +2\cdot ( 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1}) - (2n-1)\cdot 3^{-n} \\ s_n(1-3^{-1}) &=& 1 +2\cdot ( \underbrace{3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1}}_{\text{geometric progression}} ) - (2n-1)\cdot 3^{-n} \\ \dfrac{2}{3} s_n &=& 1 +2\cdot \mathbf{ S_n } - (2n-1)\cdot 3^{-n} \\ \end{array}\\ \begin{array}{|rcll|} \hline S_n &=& 3^{-1} + & 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1} \\ 3^{-1} S_n &=& & 3^{-2} + 3^{-3} + 3^{-4} + \dotsb + 3^{-n+1} + 3^{-n} \\ \hline \end{array} \\ \begin{array}{rcll} S_n-3^{-1} S_n &=& 3^{-1} - 3^{-n} \\ S_n(1-3^{-1} ) &=& 3^{-1} - 3^{-n} \\ \dfrac{2}{3} S_n &=& 3^{-1} - 3^{-n} \\ \mathbf{ S_n } &=& \mathbf{\dfrac{3}{2} ( 3^{-1} - 3^{-n} )} \\ \end{array}\)

 

\(\begin{array}{|rcll|} \hline \dfrac{2}{3} s_n &=& 1 +2\cdot \mathbf{ S_n } - (2n-1)\cdot 3^{-n} \quad | \quad \mathbf{ S_n = \dfrac{3}{2} ( 3^{-1} - 3^{-n} )} \\\\ \dfrac{2}{3} s_n &=& 1 +3\cdot ( 3^{-1} - 3^{-n} ) - (2n-1)\cdot 3^{-n} \\\\ \dfrac{2}{3} s_n &=& 1 +3^1 3^{-1} - 3^13^{-n} - (2n-1)\cdot 3^{-n} \\\\ \dfrac{2}{3} s_n &=& 1 + 1 - 3^{-n}(3 + 2n-1 ) \\\\ \dfrac{2}{3} s_n &=& 2- 3^{-n}(2n+2) \\\\ \dfrac{2}{3} s_n &=& 2- 2\cdot 3^{-n}(n+1) \\\\ \dfrac{2}{3} s_n &=& 2\Big(1- 3^{-n}(n+1)\Big) \quad | \quad : 2 \\\\ \dfrac{1}{3} s_n &=& 1- 3^{-n}(n+1) \\ s_n &=& 3\Big(1- 3^{-n}(n+1)\Big) \\ &=& 3\Big(1- \dfrac{n+1} {3^{n} }\Big) \\ &=& 3\Big(\dfrac{3^{n}-n-1} {3^{n} }\Big) \\ &=& \dfrac{3^{n}-n-1} {3^{-1}\cdot 3^{n} } \\ \mathbf{ s_n } &=& \mathbf{\dfrac{3^{n}-n-1} {3^{n-1} } } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline s = \sum \limits_{n=1}^{\infty}\dfrac{2n-1}{3^{n-1}} = \lim \limits_{n\to\infty} s_n &=& \lim \limits_{n\to\infty}\dfrac{3^{n}-n-1} {3^{n-1} } \quad | \quad \text{l'Hospital's rule} \\ &=& \lim \limits_{n\to\infty}\dfrac{n3^{n-1}-1} {(n-1)3^{n-2} } \quad | \quad \text{l'Hospital's rule} \\ &=& \lim \limits_{n\to\infty}\dfrac{n(n-1)3^{n-2}} {(n-1)(n-2)3^{n-3} } \\\\ &=& \lim \limits_{n\to\infty}\dfrac{n(n-1)3^{n-2}} {(n-1)(n-2)3^{n-2}3^{-1} } \\\\ &=& \lim \limits_{n\to\infty}\dfrac{n} { (n-2)3^{-1} } \\\\ &=& 3\cdot \lim \limits_{n\to\infty}\dfrac{n} {n-2} \\\\ &=& 3\cdot \lim \limits_{n\to\infty}\dfrac{1} {\dfrac{n-2}{n} } \\\\ &=& 3\cdot \lim \limits_{n\to\infty}\dfrac{1} {1-\dfrac{2}{n} } \quad | \quad \lim \limits_{n\to\infty} \dfrac{2}{n} = 0 \\\\ &=& 3 \cdot \dfrac{1} {1-0 } \\\\ &=& 3 \cdot \dfrac{1} {1 } \\\\ \mathbf{ s } &=& \mathbf{3} \\ \hline \end{array}\)

 

laugh

 Apr 30, 2019
 #3
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0

∑[3^(1 - k) (2 k - 1) , k, 1, 1000] = 3

 Apr 30, 2019

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