I've already shown that $\displaystyle\sum\limits_{k=2}^\infty \frac{1}{k^2 - 1} = \displaystyle\sum\limits_{k=2}^\infty \frac{1}{2(k+1)} - \frac{1}{2(k+1)}$ is convergent and tends to $\dfrac{3}{4}$.
I now must prove convergence of $\displaystyle\sum\limits_{k=1}^\infty \frac{1}{k^2} = \displaystyle\sum\limits_{k=2}^\infty \dfrac{1}{(k-1)^2}$ with this and by the help of the comparison test.
Let $a_n = \dfrac{1}{(k-1)^2} ,\; b_n =\dfrac{1}{k^2 - 1}$ but since $|a_n| \leq b_n $ is never true for any positive $k$ how I can prove that and give an upper bound?