Zeta Stuff

Calculus Level 5

∑ n = 1 ∞ ( ( − 4 π 2 ) n ( 2 ζ ′ ( − 2 n ) + n ζ ( 1 − 2 n ) + ζ ( 2 ) ) ( 2 n ) ! − 2 ) = ? \displaystyle\sum _{n=1}^{\infty } \left(\frac{\left(-4 \pi ^2\right)^n \left(2 \zeta '(-2 n)+n \zeta (1-2 n)+\zeta (2)\right)}{(2 n)!}-2\right) = \, ?


The answer is 1.

Christopher Criscitiello by Christopher Criscitiello , 25, USA

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1 solution

Here are some hints:

(1) Why can you ignore the summand ζ ( 2 ) \zeta(2) ?

(2) Start from ∑ n > 1 ζ ( n ) − 1 \displaystyle \sum_{n>1} \zeta(n)-1 . Use the functional equation for ζ \zeta . Break the sum into even and odd parts. The summands will be indeterminate. Use limits to find precise forms for each summand. Then combine the even and odd sums back together.

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