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Let \(S = \frac{1}{2^3} + \frac{1}{4^3} + \frac{1}{6^3} + \dotsb\) and \(T = \frac{1}{1^3} + \frac{1}{3^3} + \frac{1}{5^3} + \dotsb\). Find \(S/T\).

 Feb 18, 2019

Best Answer 

 #1
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+1

S=sumfor(n, 1, 1000, 1/(2*n)^3)=0.1502571129
T-sumfor(n, 1, 1000, 1/((2*n- 1 )^3))=1.05179979

S/T =0.1502571129 / 1.05179979

S/T = 1 / 7

 Feb 18, 2019
 #1
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+1
Best Answer

S=sumfor(n, 1, 1000, 1/(2*n)^3)=0.1502571129
T-sumfor(n, 1, 1000, 1/((2*n- 1 )^3))=1.05179979

S/T =0.1502571129 / 1.05179979

S/T = 1 / 7

Guest Feb 18, 2019
 #2
avatar+21848 
+4

help

Let 

\(\large{S = \dfrac{1}{2^3} + \dfrac{1}{4^3} + \dfrac{1}{6^3} + \dotsb}\)

and
\(\large{T = \dfrac{1}{1^3} + \dfrac{1}{3^3} + \dfrac{1}{5^3} + \dotsb}\).
Find

\(\large{\dfrac{S}{T}}\)

 

1.

\(\mathbf{S+T =\ ?}\)

\(\begin{array}{|rcll|} \hline S+T &=& \left(\dfrac{1}{2^3} + \dfrac{1}{4^3} + \dfrac{1}{6^3} + \dotsb \right) +\left(\dfrac{1}{1^3} + \dfrac{1}{3^3} + \dfrac{1}{5^3} + \dotsb \right) \\ S+T &=& \color{red}{\dfrac{1}{1^3} + \dfrac{1}{2^3} + \dfrac{1}{3^3} + \dfrac{1}{4^3} + \dfrac{1}{5^3} + \dfrac{1}{6^3} + \dotsb} \\ \hline \end{array}\)

 

2.

\(\mathbf{\dfrac{S}{T} =\ ?} \)

\(\begin{array}{|rcll|} \hline S &=& \dfrac{1}{2^3} + \dfrac{1}{4^3} + \dfrac{1}{6^3}+ \dfrac{1}{8^3}+ \dfrac{1}{10^3} + \dotsb \\ &=& \dfrac{1}{(2\cdot 1)^3} + \dfrac{1}{(2\cdot 2)^3} + \dfrac{1}{(2\cdot 3)^3}+ \dfrac{1}{(2\cdot 4)^3}+ \dfrac{1}{(2\cdot 5)^3} + \dotsb \\ &=& \dfrac{1}{2^3 1^3} + \dfrac{1}{2^3 2^3} + \dfrac{1}{2^3 3^3}+ \dfrac{1}{2^3 4^3}+ \dfrac{1}{2^3 5^3} + \dotsb \\ &=& \dfrac{1}{2^3} \underbrace{ \left(\color{red}{\dfrac{1}{1^3} + \dfrac{1}{2^3} + \dfrac{1}{3^3}+ \dfrac{1}{4^3}+ \dfrac{1}{5^3} + \dotsb} \right) }_{=S+T} \\ &=& \dfrac{1}{2^3}(S+T) \\\\ S &=& \dfrac{1}{8}(S+T) \quad | \quad \cdot 8 \\ 8S &=& S+T \quad | \quad -S \\ 7S &=& T \quad | \quad :7 \\ S &=& \dfrac{T}{7} \quad | \quad:T \\ \mathbf{\dfrac{S}{T}} & \mathbf{=} & \mathbf{\dfrac{1}{7}} \\ \hline \end{array}\)

 

laugh

 Feb 19, 2019

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