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Let:

$$S=\sum_{k=1}^{\infty}{k^3\over (4k^2-1)(9k^2-1)(16k^2-1)}\tag1$$

How can we show that $$S={1\over 420}+{1\over 7}\ln\left[2^{1/12}\left({4\over 3\sqrt{3}}\right)^{1/5}\right]?\tag2$$

An attempt:

Apply decomposition of partial fraction to

$${k^3\over (4k^2-1)(9k^2-1)(16k^2-1)}={A\over 4k^2-1}+{B\over 9k^2-1}+{C\over 16k^2-1}\tag3$$

$$k^3=A(9k^2-1)(16k^2-1)+B(4k^2-1)(16k^2-1)+C(4k^2-1)(9k^2-1)$$

$(1)$ becomes

$$\sum_{k=1}^{\infty}\left({32\over 15(4k^2-1)}-{32\over 35(9k^2-1)}-{43\over 21(16k^2-1)}\right)\tag4$$

Further simplify to

$${1\over 2}\sum_{k=1}^{\infty}\left[{32\over 15}\color{red}{\left({1\over 2k-1}-{1\over 2k+1}\right)}-{32\over 35}\left({1\over 3k-1}-{1\over 3k+1}\right)-{43\over 21}\left({1\over 4k-1}-{1\over 4k+1}\right)\right]\tag5$$

The red part telescope

$$\sum_{k=1}^{N}\left({1\over 2k-1}-{1\over 2k+1}\right)={1\over 2N+1}\tag6$$

So $(5)$ simplify to

$${1\over 2}\sum_{k=1}^{\infty}\left[{32\over 35}\left({1\over 3k-1}-{1\over 3k+1}\right)-{43\over 21}\left({1\over 4k-1}-{1\over 4k+1}\right)\right]\tag7$$

I got this far, but I can't evaluate any further.

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    $$\ln\left[2^{1/12}\left({4\over 3\sqrt{3}}\right)^{1/5}\right]= \frac{1}{12} \ln 2+\frac{2}{5} \ln 2-\frac{3}{10} \ln 3$$ – Yuriy S Mar 31 '17 at 14:23
  • Not sure if it affects your approach, but the partial fraction decomposition has to be with irreducible elements in the denominator... Each term factors, so you would actually have $6$ terms, not just $3$. – Clayton Mar 31 '17 at 14:33
  • Just after line 3 ... $k^3= f(k^2)$ ??? ... have you cabbaged a factor of $k$ ? ... You will use $\frac{1}{i}= \int_0^1 x^{i-1} dx $ to do those other sums. – Donald Splutterwit Mar 31 '17 at 14:39
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    $(5)$ cannot be right. Otherwise, the whole problem would boil down The whole problem boils down to the evaluation of $$ \sum_{k\geq 1}\left[\frac{1}{mk-1}-\frac{1}{mk+1}\right] = \frac{1}{m}\left[\psi\left(1+\frac{1}{m}\right)-\psi\left(1-\frac{1}{m}\right)\right]\tag{A}$$ that by the https://en.wikipedia.org/wiki/Digamma_function#Reflection_formula for the digamma function equals $$ \sum_{k\geq 1}\left[\frac{1}{mk-1}-\frac{1}{mk+1}\right]=1-\frac{\pi}{m}\cot\left(\frac{\pi}{m}\right)\tag{B}$$ and no logarithm is involved. – Jack D'Aurizio Mar 31 '17 at 15:25

1 Answers1

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By computing the residues of $f(z)=\frac{z^3}{(4z^2-1)(9z^2-1)(16z^2-1)}$ at $\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{1}{4}$ we get

$$ f(z)=\frac{1}{120}\left(\frac{1}{z-\frac{1}{2}}+\frac{1}{z+\frac{1}{2}}\right)-\frac{1}{70}\left(\frac{1}{z-\frac{1}{3}}+\frac{1}{z+\frac{1}{3}}\right)+\frac{1}{168}\left(\frac{1}{z-\frac{1}{4}}+\frac{1}{z+\frac{1}{4}}\right)\tag{A}$$ where the sum of residues is $0=\frac{1}{120}-\frac{1}{70}+\frac{1}{168}$ and $$ \sum_{n\geq 1}\left(\frac{1}{n-\frac{1}{m}}+\frac{1}{n+\frac{1}{m}}-\frac{2}{n}\right)=-2\gamma-\psi\left(1-\frac{1}{m}\right)-\psi\left(1+\frac{1}{m}\right). \tag{B}$$ It follows that the sum of the given series just depends on values of the digamma function at rational points, that can be computed through Gauss' digamma theorem. In particular: $$ \sum_{n\geq 1}\frac{n^3}{(4n^2-1)(9n^2-1)(16n^2-1)}=\frac{1}{420}\left[1+29 \log(2)-18 \log(3)\right].\tag{C}$$

Jack D'Aurizio
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