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

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

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

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 $$\eqalign{\Psi \left( 1 \right) &=-\gamma\cr \Psi \left( 5/4 \right) &=4-\gamma-3\,\ln \left( 2 \right) -\pi/2 \cr \Psi \left( 3/2 \right) &=2-\gamma-2\,\ln \left( 2 \right)\cr \Psi \left( 7/4 \right) &=4/3-\gamma-3\,\ln \left( 2 \right) +\pi/2 \cr \Psi \left( 2 \right) &=1-\gamma\cr \Psi \left( 9/4 \right) &={ {24}/{5}}-\gamma-3\,\ln \left( 2 \right) -\pi/2\cr}$$