<s>[INST] 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.
 [/INST] $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 $b$.
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