<prompt> Kokotoa sum_n0inftyfracnn4n21.  <response>: Kumbuka kuwa n4n21n42n21-n2n212-n2n2n 1n2-n1.Ikitengana katika sehemu ndogo, tunapata kwamba fracnn4n21frac12 kushotofrac1n2-n1-frac1n2n1kulia.Sasa, kumbuka kwamba ikiwa fn frac1n2-n1, kisha fn1frac1n12-n11frac 1n2n1. Inafuata kwamba sum_n0inftyfracnn4n21frac12Biglf0 -f1f1-f2f2-f3cdotsBigr.Kwa kuwa fn inaelekea 0 kama n inakuwa kubwa, jumla hii darubini hadi f02frac12. 