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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?

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  • $\begingroup$ In terms of comparison, from k=2, compare their denominators. It becomes pretty obvious then. $\endgroup$ – imranfat Nov 9 '19 at 14:58
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$$k^2-1 < k^2 \Rightarrow \frac{1}{k^2-1} > \frac{1}{k^2}\ \forall k \in \mathbb{N},\ k\geq 2$$ Therefore $$\sum\limits_{k=1}^\infty \frac{1}{k^2} = 1 + \sum\limits_{k=2}^\infty \frac{1}{k^2} < 1 + \sum\limits_{k=2}^\infty \frac{1}{k^2-1} = 1+\frac{3}{4} < \infty$$

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  • $\begingroup$ "$k^2-1 < k^2$" .. 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 $k=1$. $\endgroup$ – rndm_me Nov 9 '19 at 23:22
  • $\begingroup$ @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 $\endgroup$ – GhostAmarth Nov 10 '19 at 10:38
  • $\begingroup$ Ah, got it. And $\dfrac{\pi^2}{6} < 1+\dfrac{3}{4}$! Thanks. $\endgroup$ – rndm_me Nov 10 '19 at 15:13
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By limit comparison test the series converges indeed

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

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