<prompt> Tathminisum_m 1infty sum_n 1infty frac1mnm n 1.  <response>: Kwa sehemu ndogo,frac1mm n 1 frac1n 1 kushoto frac1m - frac1m n 1 kulia.Hivyo,anzaalignsum_m 1infty frac1mm n 1 sum_m 1infty frac1n 1 left frac1m - frac1m n 1 kulia frac1n 1 kushoto 1 - frac1n 2 kulia frac1n 1 kushoto frac11 2 - frac1n 3 kulia quad frac1n 1 kushoto frac13 - frac1n 4 kulia frac1n 1 kushoto frac14 - frac1n 5 kulia dotsb frac1n 1 kushoto 1 frac12 frac13 vidoti frac1n 1 kulia mwishoalignKwa hiyo,anzaalignsum_m 1infty sum_n 1infty frac1mnm n 1 sum_n 1infty frac 1nn 1 kushoto 1 frac12 frac13 vidoti frac1n 1 kulia sum_n 1infty frac1nn 1 sum_k 1n 1 frac1k sum_n 1infty sum_k 1n 1 frac1knn 1 sum_n 1infty kushoto frac1nn 1 sum_k 2n 1 frac1kn n 1 kulia sum_n 1infty frac1nn 1 sum_n 1infty sum_k 2n 1 frac1knn 1.endalignDarubini za kwanza za jumla kamasum_n 1infty left frac1n - frac1n 1 kulia 1.Kwa jumla ya pili, tunafanya muhtasari juu ya nambari zote chanya k na n kiasi kwamba 2 le k le n 1. Kwa maneno mengine, tunajumlisha k ge 2 na n ge k - 1, ambayo hutupatia.anzaalignjumla_k 2infty sum_n k - 1infty frac1knn 1 sum_k 2infty frac 1k sum_n k - 1infty frac1nn 1 sum_k 2infty frac1k sum_n k - 1infty kushoto frac1n - frac11 n 1 kulia sum_k 2infty frac1k cdot frac1k - 1 sum_k 2infty kushoto frac1k - 1 - frac1k kulia 1.mwishoalignKwa hiyo,sum_m 1infty sum_n 1infty frac1mnm n 1 2. 