<prompt> Kwa nambari halisi chanya x 1, kazi ya Riemann zeta zetax inafafanuliwa nazetax sum_n 1infty frac1nx.Kokotoosum_k 2infty zeta2k - 1.Kumbuka: Kwa nambari halisi x, x inaashiria sehemu ya sehemu ya x .  <response>: Kwa x ge 2,anzaalignzetax 1 frac12x frac13x dotsb le 1 frac122 frac132 dotsb 1 frac11 cdot 2 frac12 cdot 3 dotsb 1 kushoto 1 - frac12 kulia kushoto frac12 - frac13 kulia dotsb 2,endalignhivyo floor zetax floor 1. Kishazetax zetax - 1.Kwa hivyo, tunataka kujumlishasum_k 2infty zeta2k - 1 - 1 sum_k 2infty sum_n 2infty frac1101 n2k - 1.Tunabadilisha mpangilio wa majumuisho, ili kupataanzaalignsum_n 2infty sum_k 2infty frac1n2k - 1 sum_n 2infty left frac1n3 frac1n5 frac1n7 dotsb kulia sum_n 2infty frac1n31 - 1n2 sum_n 2infty frac1n3 - n.endalignKwa sehemu ndogo,frac1n3 - n frac12n - 1 - frac1n frac12n 1. Kwa hiyo,anzaalignsum_n 2infty frac1n3 - n sum_n 2infty kushoto frac12n - 1 - frac1n frac12n 1 kulia kushoto frac121 - frac12 frac123 kulia kushoto frac12 2 - frac13 frac124 kulia kushoto frac123 - frac14 frac125 kulia dotsb frac121 - frac12 frac122 frac14.mwishopanga 