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What is the sum of the following sequence: 1^-6 + 3^-6 + 5^-6 + 7^-6 + 9^-6 +.........to infinity.

Any help or hint would be greatly appreciated. Thank you.

 Sep 28, 2019

Best Answer 

 #2
avatar+23317 
+2

What is the sum of the following sequence:

\(1^{-6} + 3^{-6} + 5^{-6} + 7^{-6} + 9^{-6} + \ldots\) to infinity.

 

\(\begin{array}{|rcll|} \hline 1^{-6} + 3^{-6} + 5^{-6} + 7^{-6} + 9^{-6} + \ldots &=& 1+\dfrac{1}{3^6}+\dfrac{1}{5^6}+\dfrac{1}{7^6}+\dfrac{1}{9^6}+ \ldots \\ &=& \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^6} \\ \hline \end{array}\)

 

zeta-function:

\(\begin{array}{|rcll|} \hline \zeta(s) &=& 1+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\dfrac{1}{4^s}+\dfrac{1}{5^s}+ \ldots \\ \zeta(s)\left(1-\dfrac{1}{2^s} \right) &=& \left( 1+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\dfrac{1}{4^s}+\dfrac{1}{5^s}+ \ldots \right) \left(1-\dfrac{1}{2^s} \right) \\ &=& \left( 1+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\dfrac{1}{4^s}+\dfrac{1}{5^s}+ \ldots \right) \\ && -\left(\dfrac{1}{2^s}+\dfrac{1}{4^s}+\dfrac{1}{6^s}+\dfrac{1}{8^s} \right) \\ \mathbf{\zeta(s)\left(1-\dfrac{1}{2^s} \right) } &=& \mathbf{ 1+\dfrac{1}{3^6}+\dfrac{1}{5^6}+\dfrac{1}{7^6}+\dfrac{1}{9^6}+ \ldots } \\ \hline \end{array} \)

 

with Bernoulli numbers B:

\(\begin{array}{|lrcll|} \hline & \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^s} &=& \left(1-\dfrac{1}{2^s} \right) \zeta(s) \quad &| \quad \zeta(2n) = \Big(-1\Big)^{(n-1)} \cdot \dfrac{ (2 \pi)^{2n} } {2(2n)!} B_{2n} \\ \hline s=6: & \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^6} &=& \left(1-\dfrac{1}{2^6} \right) \zeta(6) \quad &| \quad \zeta(2\cdot 3) = \Big(-1\Big)^{(3-1)} \cdot \dfrac{ (2 \pi)^{2\cdot 3} } {2(2\cdot 3)!} B_{2\cdot 3} \\ & & & \quad &| \quad \zeta(6) = \Big(-1\Big)^{2} \cdot \dfrac{ (2 \pi)^{6} } {2(6)!} B_{6} \\ & & & \quad &| \quad \zeta(6) = \dfrac{ 2^6 \pi^6 } { 2\cdot 720} B_{6} \qquad B_6 = \dfrac{1}{42} \\ & & & \quad &| \quad \zeta(6) = \dfrac{ 2^6 \pi^6 } { 2\cdot 720} \cdot \dfrac{1}{42} \\ & & & \quad &| \quad \boxed{\zeta(6) = \dfrac{ \pi^6 } {945} } \\ & \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^6} &=& \left(1-\dfrac{1}{2^6} \right) \dfrac{ \pi^6 } {945} \\ & \mathbf{ \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^6} } &=& \mathbf{ \dfrac{ \pi^6 } {960 } } \\ \hline \end{array}\)

 

Bernoulli numbers:

 

laugh

 Oct 1, 2019
 #1
avatar+104874 
+3

   1              1            1             1            1

___   +     ___   +    ___  +    ____  +  ____ +  ........     =

1^6           3^6         5^6          7^6         9^6

 

 

∞                                           pi^6

∑   1/ [2n +1]^6       =           _____ 

n = 0                                      960

 

 

 

cool cool cool

 Sep 28, 2019
 #2
avatar+23317 
+2
Best Answer

What is the sum of the following sequence:

\(1^{-6} + 3^{-6} + 5^{-6} + 7^{-6} + 9^{-6} + \ldots\) to infinity.

 

\(\begin{array}{|rcll|} \hline 1^{-6} + 3^{-6} + 5^{-6} + 7^{-6} + 9^{-6} + \ldots &=& 1+\dfrac{1}{3^6}+\dfrac{1}{5^6}+\dfrac{1}{7^6}+\dfrac{1}{9^6}+ \ldots \\ &=& \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^6} \\ \hline \end{array}\)

 

zeta-function:

\(\begin{array}{|rcll|} \hline \zeta(s) &=& 1+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\dfrac{1}{4^s}+\dfrac{1}{5^s}+ \ldots \\ \zeta(s)\left(1-\dfrac{1}{2^s} \right) &=& \left( 1+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\dfrac{1}{4^s}+\dfrac{1}{5^s}+ \ldots \right) \left(1-\dfrac{1}{2^s} \right) \\ &=& \left( 1+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\dfrac{1}{4^s}+\dfrac{1}{5^s}+ \ldots \right) \\ && -\left(\dfrac{1}{2^s}+\dfrac{1}{4^s}+\dfrac{1}{6^s}+\dfrac{1}{8^s} \right) \\ \mathbf{\zeta(s)\left(1-\dfrac{1}{2^s} \right) } &=& \mathbf{ 1+\dfrac{1}{3^6}+\dfrac{1}{5^6}+\dfrac{1}{7^6}+\dfrac{1}{9^6}+ \ldots } \\ \hline \end{array} \)

 

with Bernoulli numbers B:

\(\begin{array}{|lrcll|} \hline & \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^s} &=& \left(1-\dfrac{1}{2^s} \right) \zeta(s) \quad &| \quad \zeta(2n) = \Big(-1\Big)^{(n-1)} \cdot \dfrac{ (2 \pi)^{2n} } {2(2n)!} B_{2n} \\ \hline s=6: & \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^6} &=& \left(1-\dfrac{1}{2^6} \right) \zeta(6) \quad &| \quad \zeta(2\cdot 3) = \Big(-1\Big)^{(3-1)} \cdot \dfrac{ (2 \pi)^{2\cdot 3} } {2(2\cdot 3)!} B_{2\cdot 3} \\ & & & \quad &| \quad \zeta(6) = \Big(-1\Big)^{2} \cdot \dfrac{ (2 \pi)^{6} } {2(6)!} B_{6} \\ & & & \quad &| \quad \zeta(6) = \dfrac{ 2^6 \pi^6 } { 2\cdot 720} B_{6} \qquad B_6 = \dfrac{1}{42} \\ & & & \quad &| \quad \zeta(6) = \dfrac{ 2^6 \pi^6 } { 2\cdot 720} \cdot \dfrac{1}{42} \\ & & & \quad &| \quad \boxed{\zeta(6) = \dfrac{ \pi^6 } {945} } \\ & \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^6} &=& \left(1-\dfrac{1}{2^6} \right) \dfrac{ \pi^6 } {945} \\ & \mathbf{ \sum \limits_{n=0}^{\infty} \dfrac{1}{(2n+1)^6} } &=& \mathbf{ \dfrac{ \pi^6 } {960 } } \\ \hline \end{array}\)

 

Bernoulli numbers:

 

laugh

heureka Oct 1, 2019
 #3
avatar+104874 
+2

Thanks, heureka.......I wondered how WolframAlpha  derived that answer....!!!

 

I want to look at this one again......

 

 

cool cool cool

 Oct 1, 2019
edited by CPhill  Oct 2, 2019
 #4
avatar+23317 
+2

Thank you, CPhill !

 

laugh

heureka  Oct 2, 2019

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