Find $\frac{1}{1.2.3.4}+\frac{4}{3.4.5.6}+\frac{9}{5.6.7.8}+\frac{16}{7.8.9.10}+\dots$

Note that after the partial fraction decomposition we have that (one of your coefficients is wrong), $$\frac{1}{24}\left(\frac{-2}{n+1}+\frac{3}{2n+1}+\frac{\color{blue}{1}}{2n-1}\right)=\frac{1}{6}\left(\frac{-1}{2n+2}+\frac{1}{2n+1}\right)+\frac{1}{24}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right).$$ Hence, as $N\to\infty$, $$\begin{align} \sum_{n=1}^{N}T_n &=\frac{1}{6}\sum_{n=1}^{N}\left(\frac{-1}{2n+2}+\frac{1}{2n+1}\right) +\frac{1}{24}\sum_{n=1}^{N}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\\ &=\frac{1}{6}\sum_{n=3}^{2N+2}\frac{(-1)^{n+1}}{n} +\frac{1}{24}\left( 1-\frac{1}{2N+1}\right)\\ &\to\frac{1}{6}\left(\log(2)-1+\frac{1}{2}\right)+\frac{1}{24}=\frac{\log(2)}{6}-\frac{1}{24} \end{align}$$ where the last series is telescopic.

P.S. WA confirms the result. Note that in ss1729's first method $$\begin{align} &S_{2n}=\frac{1}{24}\Big[-2[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{2n+1}]+3[\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{4n+1}]+[1+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{4n-1}]\Big]\\ &=\frac{1}{24}\Big[\color{red}{-2\big[\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n+1}\big]}+2\big[-\frac{1}{2}-\frac{1}{4}-\dots-\frac{1}{2n}\big]+\color{red}{2\big[\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n+1}+\frac{1}{2n+3}+\dots+\frac{1}{4n+1}\big]}+2\big[1+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n+1}+\frac{1}{2n+3}+\dots+\frac{1}{4n+1}\big]-1-\frac{2}{4n+1}\Big] \end{align}$$ and the red part does NOT go to zero!!