Find the closed form and take the limit in the case of infinite sums. That is, you find the closed form of the sum $\sum_{4 \le k \le m}\frac{1}{k^2-1}$ and evaluate $\lim_{m \to \infty}(\text{closed form of the sum})$.
In this case, you apply partial fraction decomposition to $\frac{1}{k^2-1}$ and arrive at $\frac{1}{k^2-1}=\frac{-\frac{1}{2}}{k+1}+\frac{\frac{1}{2}}{k-1}=\frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right).$
Taking the partial sum $\sum_{4 \le k \le m}\frac{1}{k^2-1}=\frac{1}{2}\sum_{4 \le k \le m}\frac{1}{k-1}-\frac{1}{k+1}.$
The sum telescopes, so we see that $\sum_{4 \le k \le m}\frac{1}{k+1}-\frac{1}{k-1}=\color{red}{\frac{1}{3}}-\color{green}{\frac{1}{5}}+\color{red}{\frac{1}{4}}-\frac{1}{6}+\color{green}{\frac{1}{5}}-\frac{1}{7}+\dots+\color{green}{\frac{1}{m-1}}-\frac{1}{m-3}+\color{red}{\frac{1}{m}}-\frac{1}{m-2}+\color{red}{\frac{1}{m+1}}-\color{green}{\frac{1}{m-1}}=\frac{1}{3}+\frac{1}{4}+\frac{1}{m}+\frac{1}{m+1}.$
Taking the limit, we have $\lim_{m \to \infty}\frac{1}{3}+\frac{1}{4}+\frac{1}{m}+\frac{1}{m+1}=\frac{1}{3}+\frac{1}{4}+\lim_{m \to \infty}\frac{1}{m}+\frac{1}{m+1}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}.$
Recalling the factor of $\frac{1}{2}$, we arrive at $\sum_{4 \le k \le \infty}\frac{1}{k^2-1}=\frac{1}{2}\cdot\frac{7}{12}=\frac{7}{24}$.