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

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?

• In terms of comparison, from k=2, compare their denominators. It becomes pretty obvious then. Nov 9, 2019 at 14:58

$$k^2-1 < k^2 \Rightarrow \frac{1}{k^2-1} > \frac{1}{k^2}\ \forall k \in \mathbb{N},\ k\geq 2$$ Therefore $$\sum\limits_{k=1}^\infty \frac{1}{k^2} = 1 + \sum\limits_{k=2}^\infty \frac{1}{k^2} < 1 + \sum\limits_{k=2}^\infty \frac{1}{k^2-1} = 1+\frac{3}{4} < \infty$$
• "$k^2-1 < k^2$" .. don't you have to modify the sums so they have the same starting index? Since $\left|\dfrac{1}{k^2}\right| \leq \dfrac{1}{k^2-1}$ would be true for k > 1. In my definition the two series have both start at $k=1$. Nov 9, 2019 at 23:22
• @rndm_me That’s why we pull the first term out of the sum. Then we have two series starting at $k=2$ and we know that one is strictly smaller than the other one and therefore finite Nov 10, 2019 at 10:38
• Ah, got it. And $\dfrac{\pi^2}{6} < 1+\dfrac{3}{4}$! Thanks. Nov 10, 2019 at 15:13
$$\frac{\frac{1}{k^2 }}{\frac{1}{k^2 - 1}}=\frac{k^2 - 1}{k^2} \to 1$$