Dtyjlui

I am investigating (just for fun) the sum
$$S=sum_{n=2}^{infty}frac1{n^3-1}$$
Wolfram Alpha gives me the 'value'
$$S=-frac13sum_{{omega,:,omega^3+6omega^2+12omega+7=0}}frac{psi_{0}(-omega)}{omega^2+4omega+4}$$
Where $psi_{0}(s)$ is the di-gamma function. I would like to know how this is found.

My attempts: recall that $$x^3-1=prod_{k=1}^{3}(x-alpha_k)$$
Where $alpha_k=expfrac{2ipi(k-1)}{3}$. Hence
$$S=sum_{ngeq2}prod_{k=1}^3frac1{n-alpha_k}$$
I have shown in other posts that given some non-zero sequence ${a_k:k=1,2,..,m}$ where $jneq kiff a_jneq a_k$ then
$$prod_{k=1}^{m}frac1{x-a_k}=sum_{k=1}^{m}frac1{x-a_k}prod_{j=1\jneq k}^{m}frac1{a_k-a_j}$$
So
$$S=sum_{ngeq2}bigg[sum_{k=1}^{3}frac1{n-alpha_k}prod_{j=1\jneq k}^{3}frac1{alpha_k-alpha_j}bigg]$$
Setting $b(k)=prod_{j=1\jneq k}^{3}frac1{alpha_k-alpha_j}$,
$$S=sum_{ngeq2}sum_{k=1}^3frac{b(k)}{n-alpha_k}$$
$$S=b(1)sum_{ngeq2}frac1{n-alpha_1}+b(2)sum_{ngeq2}frac1{n-alpha_2}+b(3)sum_{ngeq2}frac1{n-alpha_3}$$
Then focusing on
$$begin{align}
S_k=&sum_{ngeq2}frac1{n-alpha_k}\
=&sum_{ngeq2}int_0^1x^{n-1}frac{mathrm dx}{x^{alpha_k}}\
=&int_0^1frac{1}{x^{alpha_k-1}}sum_{ngeq0}x^{n}mathrm dx\
=&int_0^1x^{1-alpha_k}(1-x)^{-1}mathrm dx\
=&text{???}
end{align}$$
for this final integral I considered using the beta function but that wouldn't work because $Gamma(0)$ is undefined. Plus I'm not sure if $S_k$ even converges.

As you can see, I'm stuck. Could I have a little help?

Thanks

Edit:

It can be seen here that there seems to be a pattern to evaluations of the function $$S(k)=sum_{ngeq2}frac1{n^{2k+1}-1},qquad kinBbb N$$
Which seem to be in the form
$$S(k)=-frac1{1+2k}sum_{{omega,:,R_k(omega)=0}}frac{psi_0(-omega)}{P_k(omega)}$$
Where $R_k(x)$ is a polynomial of degree $k$ and $P_k(x)$ is a polynomial of degree $k-1$.

Partial fraction decomposition tells us that

$$frac{1}{n^k-1}=sum_{omega^k=1} frac{1}{n-omega}lim_{ztoomega}frac{z-omega}{z^k-1}overset{text{L'H}}{=}frac{1}{k}sum_{omega^k=1} frac{omega}{n-omega}$$

and by the property that

$$psi(z+1)=psi(z)+frac{1}{z}$$

we see that

$$frac{1}{n^k-1}=frac{1}{k}sum_{omega^k=1} omegaBig(psi(n-omega+1)-psi(n-omega)Big)$$

Then we may sum both sides from 2 to infinity:

$$sum_{n=2}^{infty}frac{1}{n^k-1}=frac{1}{k} lim_{h to infty}sum_{omega^k=1} omegaBig(psi(h-omega)-psi(2-omega)Big)=-frac{1}{k}sum_{omega^k=1} omega ,psi(2-omega)$$

where the second equality is because $psi(h)=log(h)+O(1/h)$.

Arriving at the WolframAlpha result:

Plug in $k=3$, and thus

$$begin{align*}&quadsum_{n=2}^{infty}frac{1}{n^3-1}\
&=-frac{1}{3}sum_{omega^3=1}omega,psi(2-omega)\
&=-frac{1}{3}sum_{omega^3=1}frac{psi(2-omega)}{omega^2}quadbig(omega^3=1big)\
&=-frac{1}{3}sum_{(omega+2)^3=1}frac{psi(-omega)}{(omega+2)^2}quadbig(omegamapstoomega+2big)\
&=-frac{1}{3}sum_{omega^3+6omega^2+12omega+7=0}frac{psi(-omega)}{omega^2+4omega+4}
end{align*}$$

which is what we wanted.

$$sum_{ngeq 0}frac{1}{(n+a)(n+b)}=frac{psi(a)-psi(b)}{a-b} $$
leads by partial fraction decomposition to
$$sum_{ngeq 0}frac{1}{(n+a)(n+b)(n+c)} = frac{1}{(a-b)(b-c)(c-a)}left[(b-c)psi(a)+(c-a)psi(b)+(a-b)psi(c)right] $$
In your case $a,b,c$ are the opposite of the roots of $(x+2)^3-1$, and I'll let you finish the computations: $psi(1)=-gamma$, but $frac{1}{2}(5pm isqrt{3})$ are not special arguments for $psi(x)=frac{d}{dx}logGamma(x)$.