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Let $\zeta$ be the Riemann- Zeta function. For any integer, $n \geq 2$, how to prove $$\zeta(2) \zeta(2n-2) + \zeta(4)\zeta(2n-4) + \cdots + \zeta(2n-2)\zeta(2) = \Bigl(n + \frac{1}{2}\Bigr)\zeta(2n)$$

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It would help a great deal to change all your zetas to Bernoulli numbers. – J. M. Oct 23 '10 at 13:14

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up vote 13 down vote accepted

This is too nice an exercise to give away. Your starting point should be: $$\frac{t}{e^t-1} = 1 - t/2 + \frac{2 \zeta(2)}{(2 \pi)^2} t^2 - \frac{2 \zeta(4)}{(2 \pi)^4} t^4 + \frac{2 \zeta(6)}{(2 \pi)^6} t^6 - \cdots.$$

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Wow, expressed this way I hardly can recognize that power series... :) – J. M. Oct 23 '10 at 14:33

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