Finding the sum of a series $\sum_{n=4}^{\infty}\frac{1}{n^{2}-1}$ I was thinking about this problem:
Given a series $\sum_{n=4}^{\infty}\frac{1}{n^{2}-1}$ ,how can i show that its sum is a/an rational/irrational number,given that the series converges?
Could someone point me in the right direction? Thanks everyone in advance.

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 $\sum_{4 \le k \le \infty}\frac{1}{k^2-1}=\frac{1}{2}\cdot\frac{7}{12}=\frac{7}{24}$.
