Convergence of $\sum_{k=1}^\infty \frac{1}{k^2}$ by using $\sum_{k=2}^\infty \frac{1}{k^2 - 1}$ and comparison test

Question

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?

Answer 1

$$k^2-1 < k^2 \Rightarrow \frac{1}{k^2-1} > \frac{1}{k^2}\ \forall k \in \mathbb{N},\ k\geq 2$$ Therefore $$\sum\limits_{k=1}^\infty \frac{1}{k^2} = 1 + \sum\limits_{k=2}^\infty \frac{1}{k^2} < 1 + \sum\limits_{k=2}^\infty \frac{1}{k^2-1} = 1+\frac{3}{4} < \infty$$

Answer 2

By limit comparison test the series converges indeed

$$\frac{\frac{1}{k^2 }}{\frac{1}{k^2 - 1}}=\frac{k^2 - 1}{k^2} \to 1$$