What does this sum $\sum_{t=c}^{n-1}(\frac{1}{t^2-1})$ equal?
For context, I am trying to digest the cardinal payoff variant of the secretary problem. There is an interview process with $n$ candidates each having some value associated with them which comes from i.i.d. interval $[0;1]$, the relative value of candidates is not known, only thing learned when interviewing a candidate is if the candidate is best so far. On the spot the interviewer must evaluate whether or not to accept the candidate and this decision can not be reverted. If none are chosen, the last one must be taken. The goal is to find a strategy which maximizes the value of the chosen candidate.
To my understanding between different strategies, the optimal strategy involves some cutoff rule, where the first $c-1$ candidates are immediately rejected and the next best one is chosen. The value function in this case can be expressed iteratively by the formula: $V_n(t) = \frac{1}{t}\cdot E_t + (1- \frac{1}{t})\cdot V_n(t+1)$ which is the same as $V_n(t) = \frac{1}{t+1} + (1 - \frac{1}{t})\cdot V_n(t+1)$, since $E_t = \frac{t}{t+1}$ - the expected value of a canditate $t$ when he is the best of the first $t$ candidates. This however can be written as an iterated binary operation for a cutoff value $c$:
$V_n(c)= \sum_{t=c}^{n-1}[\Pi_{s=c}^{t-1}(\frac{s-1}{s})]\cdot (\frac{1}{t+1}) + [\Pi_{s=c}^{n-1}(\frac{s-1}{s})]\cdot\frac{1}{2} $
(^ This can also be found on the wiki page, also in this paper)
However when evaluating the iterated binary operations I stumble into problems. I get the trick for the iterated multiplication, the cross division is quite interesting, for example: $\Pi_{s=c}^{n-1}(\frac{s-1}{s}) = \frac{c-1}{n-1}$ since the other terms just cross out when writen out. So now the equation becomes
$[(c-1)\cdot\sum_{t=c}^{n-1}(\frac{1}{t-1})\cdot(\frac{1}{t+1})] + \frac{c-1}{n-1}\cdot \frac{1}{2}$
since $(c-1)$ is a constant in each sum iteration, it can be factored out. Now, how does one evaluate $\sum_{t=c}^{n-1}(\frac{1}{t-1})\cdot(\frac{1}{t+1})$? I understand that this equals $\sum_{t=c}^{n-1}(\frac{1}{t^2-1})$, but I still do not know how to evaluate it. Any help would be appreciated.
$$ \begin{align} \sum_{t=c}^{n-1}\frac1{t^2-1} &=\frac12\sum_{t=c}^{n-1}\left(\frac1{t-1}-\frac1{t+1}\right)\tag{1a}\\ &=\frac12\left(\frac1{c-1}+\frac1c-\frac1{n-1}-\frac1n\right)\tag{1b} \end{align} $$ Explanation:
$\text{(1a):}$ partial fractions
$\text{(1b):}$ telescoping series