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For a positive real number \(x > 1,\) the Riemann zeta function \(\zeta(x)\) is defined by \(\zeta(x) = \sum_{n = 1}^\infty \frac{1}{n^x}.\)
Compute \(\sum_{k = 2}^\infty \{\zeta(2k - 1)\}.\)
Note: For a real number \(x,\)\(\{x\}\) denotes the fractional part of \(x.\)

 Mar 26, 2019
 #1
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The answer is 1/4

 

If you reply I might explain why.

 Mar 26, 2019
 #2
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How did you get that?

Guest Mar 26, 2019
 #3
avatar+5651 
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\(\text{One key part is that for }k\geq 2,~\{\zeta(k)\}=\zeta(k)-1\)

 

\(\text{That applies to all of the terms of the series we are to compute}\)

 

\(\text{What you do is this....}\\ \text{Consider the sum of the fractional parts of the zeta functions for all integers 2 and larger}\\ \text{Write out horizontally some of the terms of some of each series in the sum, for a few terms, i.e. }\\ \{\zeta(2)\} = \zeta(2)-1 = (1/2)^2 + (1/3)^2 + (1/4)^2 + \dots\\ \{\zeta(3)\} = \zeta(3)-1 = (1/2)^3 + (1/3)^3 + (1/4)^3 + \dots\\ \{\zeta(4)\} = \zeta(4)-1 = (1/2)^4 + (1/3)^4 + (1/4)^4 + \dots\\ \vdots\)

 

\(\text{Now consider the sums vertically for each term}\\ \sum \limits_{k=2}^\infty (1/2)^k = \dfrac 1 2\\ \vdots\\ \sum \limits_{k=2}^\infty~(1/n)^k = \dfrac{1}{n(n-1)}\)

 

\(\text{And then you take the sum over these simplified terms}\\ \sum \limits_{n=2}^\infty \dfrac{1}{n(n-1)} = 1\)

 

\(\text{I leave it to you to modify this whole thing slightly to just sum the odd terms}\\ \text{greater than 2, as is asked for in the problem}\\ \text{You'll find the odd terms } > 2 \text{ sum to }\dfrac 1 4\\ \text{and the even terms }\geq 2 \text{ sum to }\dfrac 3 4\)

Rom  Mar 27, 2019
edited by Rom  Mar 27, 2019

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