Show $\sum_{m=1}^\infty \left( \zeta(2m) - \zeta(2m + 2k) \right) = \sum_{m=1}^k \zeta(2m) - k$

Question

I noticed this pattern using WolframAlpha that

$$ \sum_{m=1}^\infty \left( \zeta(2m) - \zeta(2m + 2k) \right) = \sum_{m=1}^k \zeta(2m) - k $$

I've tried to prove it as follows:

\begin{align} \sum_{m=1}^\infty \left( \zeta(2m) - \zeta(2m + 2k) \right) &= \sum_{m=1}^\infty \sum_{n=2}^\infty \left( \frac{1}{n^{2m}} - \frac{1}{n^{2m + 2k}} \right) \\ &= \sum_{n=2}^\infty \sum_{m=1}^\infty \left( \frac{1}{n^{2m}} - \frac{1}{n^{2m + 2k}} \right) \\ &= \sum_{n=2}^\infty \left( \sum_{m=1}^\infty \frac{1}{n^{2m}} - \sum_{m=1}^\infty \frac{1}{n^{2m + 2k}} \right) \\ &= \sum_{n=2}^\infty \left( \frac{1}{n^2 - 1} - \frac{1}{n^{2k}(n^2 - 1)} \right) \\ &= \sum_{n=2}^\infty\frac{n^{2k} - 1}{n^{2k}(n^2 - 1)} \\ &= \sum_{m=1}^k \zeta(2m) - k \end{align}

but, to complete the proof, I'm curious how WolframAlpha determines that

$$\sum_{n=2}^\infty\frac{n^{2k} - 1}{n^{2k}(n^2 - 1)} = \sum_{m=1}^k \zeta(2m) - k$$

Update:

Based on the comments, we have

$$\sum_{m=1}^\infty \left( \zeta(2m) - \zeta(2m + 2k) \right) = \sum_{n=2}^\infty\frac{n^{2k} - 1}{n^{2k}(n^2 - 1)} = \sum_{n=2}^\infty\frac{1 - n^{-2k}}{n^2 - 1} \tag{1}$$

and

$$\sum_{m=1}^k \zeta(2m) = \sum_{m=1}^k \sum_{n=1}^\infty \frac{1}{n^{2m}} = \sum_{n=1}^\infty \sum_{m=1}^k \frac{1}{n^{2m}}$$

For $n=1$,

$$\sum_{m=1}^k \frac{1}{1^{2m}} = k$$

For $n \ge 2$,

$$\sum_{m=1}^k \frac{1}{n^{2m}} = \frac{1 - n^{-2k}}{n^2 - 1}$$

Therefore,

$$\sum_{m=1}^k \zeta(2m) = k + \sum_{n=2}^\infty \frac{1 - n^{-2k}}{n^2 - 1}$$

So,

$$\sum_{m=1}^k \zeta(2m) - k = \sum_{n=2}^\infty \frac{1 - n^{-2k}}{n^2 - 1} \tag{2}$$

Finally, comparing $(1)$ and $(2)$:

$$ \sum_{m=1}^\infty \left( \zeta(2m) - \zeta(2m + 2k) \right) = \sum_{m=1}^k \zeta(2m) - k $$

References:

Show $\sum_{m=1}^{\infty} \left(\zeta(2m) - \zeta(2m+1)\right) = \frac{1}{2}$

Answer 1

If $\lim\limits_{m\to\infty}a_m=0$ then $\sum_{m=1}^\infty(a_m-a_{m+k})=\sum_{m=1}^k a_m$ (for a proof, consider finite sums...).

Now apply this to $a_m=\zeta(2m)-1$ (so, riemann-zeta is basically a distraction here).