Formula for the sum $\sum_{i=2}^{n} \frac1{i^2-1}$
Solution 1:
$$\frac1{k^2-1} = \frac1{2} \left( \frac1{k-1} - \frac1{k+1} \right)$$ Now let the telescopic summation take over.
Formula for the sum $\sum_{i=2}^{n} \frac1{i^2-1}$
Solution 1:
$$\frac1{k^2-1} = \frac1{2} \left( \frac1{k-1} - \frac1{k+1} \right)$$ Now let the telescopic summation take over.