Proving that $\sum_{k=1}^{\infty} \frac{3408 k^2+1974 k-720}{128 k^6+480 k^5+680 k^4+450 k^3+137 k^2+15 k} = \pi$

I am trying to prove that $$\sum_{k=1}^{\infty} \frac{3408 k^2+1974 k-720}{128 k^6+480 k^5+680 k^4+450 k^3+137 k^2+15 k} = \pi$$

This is what I’ve tried to simplify the sum:

$$\frac{3408 k^2+1974 k-720}{128 k^6+480 k^5+680 k^4+450 k^3+137 k^2+15 k} = 0 \implies k = \frac{\sqrt{380881}-329}{1136}
$$

The minimal polynomial of the root is $568 k^2+329 k-120$.

Now, seeing as the numberator also has the form $xk^2+yk+z$, and $\frac{720}{120} = 6,\,\frac{1974}{329} = 6,\,\frac{3408}{568} = 6$; I tried to reduce the fraction by dividing by $6$, which yielded:
$$\frac{1}{6}\frac{3408 k^2+1974 k-720}{128 k^6+480 k^5+680 k^4+450 k^3+137 k^2+15 k} = \frac{568 k^2+329 k-120}{128 k^6+480 k^5+680 k^4+450 k^3+137 k^2+15 k}$$

which implies that $$\sum_{k=1}^{\infty} \frac{568 k^2+329 k-120}{128 k^6+480 k^5+680 k^4+450 k^3+137 k^2+15 k} = \frac{\pi}{6}$$

Where to go from here?

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$$
\sum_{k=1}^{\infty}\frac{3408 k^2+1974 k-720}{128 k^6+480 k^5+680 k^4+450 k^3+137 k^2+15 k}
$$
$$=\sum_{k=1}^{\infty}
\left(
-\frac{192}{4k}+\frac{1334}{4k+1}-\frac{2280}{4k+2}+\frac{756}{4k+3}+\frac{952}{4k+4}-\frac{570}{4k+5}\right)
$$

$$
=
-952 \color{blue}{\underbrace{
\sum_{k=1}^{\infty}\left( \frac{1}{4k}-\frac{1}{4k+4} \right)}_{A}}
+ 570 \color{darkgreen}{\underbrace{
\sum_{k=1}^{\infty}\left( \frac{1}{4k+1}-\frac{1}{4k+5} \right)}_{B}}
+760 \color{red}{\underbrace{\sum_{k=1}^{\infty} \left( \frac{1}{4k}+\frac{1}{4k+1}-\frac{3}{4k+2}+\frac{1}{4k+3} \right)}_{C}}
+4 \color{darkviolet}{\underbrace{
\sum_{k=1}^{\infty} \left( \frac{1}{4k+1}-\frac{1}{4k+3} \right)}_{D}}
$$
$$
=
-952\cdot \color{blue}{\frac{1}{4}}
+570\cdot \color{darkgreen}{\frac{1}{5}}
+760\cdot \color{red}{\frac{1}{6}}
+4 \color{darkviolet}{\left(\frac{\pi}{4}-\frac{2}{3}\right)}=\Large{\pi}.
$$


Small explanation:

$$
\color{blue}{A=} \frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\ldots \color{blue}{=\frac{1}{4}}.
$$
$$
\color{darkgreen}{B=} \frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\ldots \color{darkgreen}{=\frac{1}{5}}.
$$

$$
\color{red}{C=}
\sum_{k=1}^{\infty}
\left( -\frac{1}{4k}+\frac{1}{4k+1}-\frac{1}{4k+2}+\frac{1}{4k+3} \right) +
\sum_{k=1}^{\infty}
\left( \frac{2}{4k}-\frac{2}{4k+2}\right)
$$
$$
=
\sum_{n=4}^{\infty}\frac{(-1)^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{(-1)^n}{n} =
\Bigl(\ln 2 – 1 +\frac{1}{2}-\frac{1}{3}\Bigr) – \Bigl(\ln 2-1\Bigr) \color{red}{=\frac{1}{2} – \frac{1}{3} = \frac{1}{6}}.
$$

$$
\color{darkviolet}{D=}
\sum_{n=2}^{\infty} \frac{(-1)^{n}}{2n+1} = \frac{\pi}{4} – 1 + \frac{1}{3} \color{darkviolet}{=\frac{\pi}{4} – \frac{2}{3}}.
$$

Here we used $\ln 2$ series (for $C$) and Leibniz formula (for $D$).


Same way we can generate other similar series for $\large \pi$:

$$
0\cdot A + 0\cdot B + 8 \cdot C + 4 \cdot D =
\sum_{k=1}^{\infty}
\frac
{80k^2+88k+12}
{k(2k+1)(4k+1)(4k+3)}
= \Large \pi.
$$
$$
-40\cdot A + 0\cdot B + 76 \cdot C + 4 \cdot D
=
\sum_{k=1}^{\infty}
\frac
{228k^2+225k+27}
{k(k+1)(2k+1)(4k+1)(4k+3)}
= \Large \pi.
$$

Exclusivity of your series is better asymptotic: ~ $\dfrac{1}{k^4}$.

Note that for $a > 0$ and $b \ge 0$,
$$\sum_{k=1}^N \dfrac{1}{ak+b} = \frac{\Psi(N+1+b/a) – \Psi(1+b/a)}{a}
= \frac{\ln(N) – \Psi(1+b/a) + O(1/N)}{a}$$
where $\Psi$ is the digamma function. It may help to know
$$\eqalign{\Psi \left( 1 \right) &=-\gamma\cr
\Psi \left( 5/4 \right) &=4-\gamma-3\,\ln \left( 2 \right) -\pi/2 \cr
\Psi \left( 3/2 \right) &=2-\gamma-2\,\ln \left( 2 \right)\cr
\Psi \left( 7/4 \right) &=4/3-\gamma-3\,\ln \left( 2 \right) +\pi/2 \cr
\Psi \left( 2 \right) &=1-\gamma\cr
\Psi \left( 9/4 \right) &={ {24}/{5}}-\gamma-3\,\ln \left( 2
\right) -\pi/2\cr}$$