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Can anyone solve this?

 

Is there a formula for this?

 Mar 13, 2019
 #1
avatar+6046 
+2

\(\dfrac{\frac 7 8 \zeta(3)}{\zeta(3)} = \dfrac 7 8\)

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 Mar 13, 2019
 #2
avatar+929 
+1

How did you get that? And what is that notation?

dgfgrafgdfge111  Mar 14, 2019
 #3
avatar+6046 
+2

That came out of Mathematica.

 

The squiggle stands for the Riemann Zeta function.

 

It's really more advanced stuff than I know.

Rom  Mar 14, 2019
 #4
avatar+6046 
+2

\(\text{I have a simpler answer.}\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}+\dfrac{1}{(2k)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}}=1\)

 

\(\sum \limits_{k=1}^\infty \dfrac{1}{(2k)^3} = \sum \limits_{k=1}^\infty \dfrac{1}{8k^3} = \dfrac 1 8 \sum \limits_{k=1}^\infty \dfrac{1}{k^3}\)

 

\(\dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}+\dfrac 1 8 \sum \limits_{k=1}^\infty\dfrac{1}{k^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}}=1\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}} + \dfrac 1 8 = 1\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}} = \dfrac 7 8\)

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 Mar 14, 2019
edited by Rom  Mar 14, 2019
 #5
avatar+929 
0

Thanks for the help!!!

dgfgrafgdfge111  Mar 14, 2019

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