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.
 A: $1/(n-1)-1/(n+1) = 2/(n^2-1)
$, so
$$\sum_{n=4}^{\infty} \frac{1}{n^2-1}
= (1/2)\sum_{n=4}^{\infty} \left(\frac{1}{n-1} - \frac{1}{n+1}\right)
= \frac{1}{3} + \frac{1}{4}
$$
since all the later terms are cancelled out.
Whoops - as pointed out by Limitless, this should be
$\frac{1}{2}\left(\frac{1}{3} + \frac{1}{4}\right)$.
Note that this also allows you to get an explicit expression for
$\sum_{n=a}^b \frac{1}{n^2-1}$
for any integers $a$ and $b$.
A: 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 $\sum_{4 \le k \le \infty}\frac{1}{k^2-1}=\frac{1}{2}\cdot\frac{7}{12}=\frac{7}{24}$.
