Closed form of $ \sum_{n=0}^{\infty} \frac{3072n^2 + 3072n + 832}{4096n^6 + 12288n^5 + 14592n^4 + 8704n^3 + 2736n^2 + 432n + 27} $

Question

I ran into the series

$$ \sum_{n=0}^{\infty} \frac{3072n^2 + 3072n + 832}{4096n^6 + 12288n^5 + 14592n^4 + 8704n^3 + 2736n^2 + 432n + 27} $$

and I’m trying to figure out one things:

1. Is there a nice closed‐form for its sum?

What I’ve tried so far:

For large $n$, the term behaves like

$$ \frac{3072n^2}{4096n^6} \;=\;\tfrac34\,n^{-4}, $$

so since $\sum n^{-4}$ converges, I’m pretty sure the original series converges absolutely.

I thought maybe the denominator factors in a helpful way (e.g.\ $(4n+\alpha)^k$) or that one could write

$$ \frac{3072n^2 + \cdots}{\text{(denominator)}} = A\bigl[f(n)-f(n+1)\bigr], $$

but I haven’t been able to match coefficients or find a telescoping pattern.

I briefly considered expressing the sum in terms of polylogarithms or known zeta‐function values, but didn’t find any standard identity that fits.

Does anyone recognize a trick to turn this into a telescoping sum, or a contour‐integral / root‐of‐unity approach, or something else that gives a closed form? Even a reference to a similar example would be really helpful. Thanks in advance!

Answer 1

$$S= \sum_{n=0}^{\infty} \frac{3072n^2 + 3072n + 832}{4096n^6 + 12288n^5 + 14592n^4 + 8704n^3 + 2736n^2 + 432n + 27} $$

Notice the first term of denominator, $4096n^6$ is $64n^3\times 64n^3$, that was a hint to the factoring of $4n+\text{something}$; $$4096n^6 + 12288n^5 + 14592n^4 + 8704n^3 + 2736n^2 + 432n + 27=(4n+1)^3(4n+3)^3$$

Using partial fractions, $$\frac{3072n^2 + 3072n + 832}{4096n^6 + 12288n^5 + 14592n^4 + 8704n^3 + 2736n^2 + 432n + 27}=\frac{32}{(4n+1)^3}-\frac{32}{(4n+3)^3}$$

$$S=32\sum_{n=0}^{\infty}\frac{1}{(4n+1)^3}-32\sum_{n=0}^{\infty}\frac{1}{(4n+3)^3}=32S_1-32S_2$$

Using the definition of Hurzwitz Zeta function, $$\color{blue}{\displaystyle \zeta (s,a)=\sum _{n=0}^{\infty }{\frac {1}{(n+a)^{s}}}}$$

$$S_1=\frac{1}{64}\sum_{n=0}^{\infty}\frac{1}{\left(n+\frac14\right)^3}=\frac{1}{64}\zeta \left(3,\frac14\right)$$

$$S_2=\frac{1}{64}\sum_{n=0}^{\infty}\frac{1}{\left(n+\frac34\right)^3}=\frac{1}{64}\zeta \left(3,\frac34\right)$$

$$\therefore S=\frac{1}{2}\zeta \left(3,\frac14\right)-\frac{1}{2}\zeta \left(3,\frac34\right)=\pi^3$$

An interesting identity,

$$\text{Cl}_m\left(\frac{\pi}2\right) = \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}2\right)}{k^m}=\sum_{n=0}^\infty\left(\frac1{(4n+1)^m}-\frac1{(4n+3)^m}\right)$$

$$\text{Cl}_3\left(\frac{\pi}2\right) = \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}2\right)}{k^3}=\sum_{n=0}^\infty\left(\frac1{(4n+1)^3}-\frac1{(4n+3)^3}\right)$$

$$\therefore S=32\text{Cl}_3\left(\frac{\pi}2\right)$$

Answer 2

Doing the same as @Amrut Ayan,

$$S_p=\sum_{n=0}^p \Bigg(\frac{32}{(4n+1)^3}-\frac{32}{(4n+3)^3}\Bigg)$$ $$S_p=\frac{1}{4} \left(\psi ^{(2)}\left(p+\frac{5}{4}\right)-\psi ^{(2)}\left(p+\frac{7}{4}\right)+\psi ^{(2)}\left(\frac{3}{4}\right)-\psi ^{(2)}\left(\frac{1}{4}\right)\right)$$

Asymptotically $$S_p=\frac{1}{4} \left(\psi ^{(2)}\left(\frac{3}{4}\right)-\psi ^{(2)}\left(\frac{1}{4}\right)\right)-\frac{1}{4 p^3}+\frac{3}{4 p^4}+O\left(\frac{1}{p^5}\right)$$ and $$\frac{1}{4} \left(\psi ^{(2)}\left(\frac{3}{4}\right)-\psi ^{(2)}\left(\frac{1}{4}\right)\right)=\pi ^3$$ Using the truncated series, the relative error is $3.21\times 10^{-3}$ percent if $p=4$.

Answer 3

Knowing that these values come from the Hurwitz zeta function, we can use the Euler-Maclaurin formula. Let $f(n)$ be the summand. By the Euler–Maclaurin formula, $$S=\frac{32}{9}+\frac{832}{54}+\int_0^{+\infty}f^{(1)}(t)\tilde B_1(t)\ {\rm d}t,\quad f^{(1)}(t) = -6144\frac{(2t+1)(16t^2+16t+5)}{(16t^2+16t+3)^4}.$$ Now the Euler–Maclaurin formula on $\zeta(3,\tfrac{1}{4})$ and $ \zeta(3,\tfrac{3}{4})$ yields, $$\begin{align*}\zeta(3,\tfrac{1}{4})=\sum_{n=0}^\infty \frac{1}{(n+1/4)^3}=8+32+\int_0^{+\infty} g_1^{(1)}(t)\tilde B_1(t)\ {\rm d}t&,\quad g^{(1)}_1(t) = -\frac{3}{(t+1/4)^4},\\ \zeta(3,\tfrac{3}{4})=\sum_{n=0}^\infty\frac{1}{(n+3/4)^3} = \frac{8}{9}+\frac{32}{27}+\int_0^{+\infty}g_2^{(1)}(t)\tilde B_1(t)\ {\rm d}t &,\quad g_2^{(1)}(t) = -\frac{3}{(t+3/4)^4}.\end{align*}$$ One can check $\frac{1}{2}(g_1^{(1)}(t) - g_2^{(1)}(t)) = f^{(1)}(t)$, so it turns out that, $$\frac{1}{2}\zeta(3,\tfrac{1}{4})-\frac{1}{2}\zeta(3,\tfrac{3}{4}) = \frac{32}{9}+\frac{832}{54}+\int_0^{+\infty} f^{(1)}(t)\tilde B_1(t)\ {\rm d}t=S.$$