<prompt> Kokotasum_n 2infty frac4n3 - n2 - n 1n6 - n5 n4 - n3 n2 - n .  <response>: Kwanza, tunatenganisha frac4n3 - n2 - n 1n6 - n5 n4 - n3 n2 - n katika sehemu ndogo. Tunazingatia dhehebu:anzaalignn6 - n5 n4 - n3 n2 - n nn5 - n4 n3 - n2 n - 1 nn4 n - 1 n2 n - 1 n - 1 nn - 1n4 n2 1 nn - 1n4 2n2 1 - n2 nn - 1n2 12 - n2 nn - 1n2 n 1n2 - n 1.endalignKisha kwa sehemu ndogo,frac4n3 - n2 - n 1nn - 1n2 n 1n2 - n 1 fracA n fracBn - 1 fracCn Dn2 n 1 fracEn Fn2 - n 1kwa baadhi ya vipengele A,B,C,D,E, na F.Kuzidisha pande zote mbili kwa nn - 1n2 n 1n2 - n 1, tunapataanzaalign4n3 - n2 - n 1 An - 1n2 n 1n2 - n 1 quad Bnn2 n 1n2 - n 1 quad Cn Dnn - 1n2 - n 1 quad En Fnn - 1n2 n 1.endalignKuweka n 0, tunapata -A 1, hivyo A -1.Kuweka n 1, tunapata 3B 3, hivyo B 1. Mlinganyo ulio hapo juu unakuwaanzaalign4n3 - n2 - n 1 -n - 1n2 n 1n2 - n 1 quad nn2 n 1n2 - n 1 quad Cn Dnn - 1n2 - n 1 quad En Fnn - 1n2 n 1.endalignHii hurahisishan4 4n3 - 2n2 - n Cn Dnn - 1n2 - n 1 En Fnn - 1n 2 n 1.Kugawanya pande zote mbili kwa nn - 1, tunapata-n2 3n 1 Cn Dn2 - n 1 En Fn2 n 1.Kupanua, tunapata-n2 3n 1 C En3 C D E Fn2 C - D E Fn D F. Kulinganisha coefficients, tunapataanzaalignC E 0, -C D E F -1, C - D E F 3, D F 1.mwishoalignKwa kuwa C E 0, -D F 3. Hivyo, D -1 na F 2. Kisha -C E -2, hivyo C 1 na E -1. Kwa hiyo,frac4n3 - n2 - n 1n6 - n5 n4 - n3 n2 - n frac1n - 1 - frac1n fracn - 1n2 n 1 - fracn - 2n2 - n 1.Kishaanzaalignjumla_n 2infty frac4n3 - n2 - n 1n6 - n5 n4 - n3 n2 - n kushoto 1 - frac12 frac17 kulia quad kushoto frac12 - frac13 frac213 - frac17 kulia quad kushoto frac13 - frac14 frac321 - frac213 kulia dotsb 1.mwishopanga 