<prompt> Kwa nambari kamili n, letH_n 1 frac12 frac13 dots frac1n.Kokotoosum_n 1infty frac1n 1 H_n H_n 1.  <response>: Tunaweza kuandikafrac1n 1 H_n H_n 1 fracfrac1n 1H_n H_n 1 fracH_ n 1 - H_nH_n H_n 1 frac1H_n - frac1H_n 1.Hivyo,anzaalignsum_n 1infty frac1n 1 H_n H_n 1 sum_n 1infty left frac11 H_n - frac1H_n 1 kulia kushoto frac1H_1 - frac1H_2 kulia kushoto frac1H_2 - frac1H_3 kulia kushoto frac1H_3 - frac1H_4 kulia dotsb frac1H_1 1.endalignKumbuka kwamba matokeo haya yanategemea ukweli kwamba H_n to infty kama n to infty. Tunaweza kuthibitisha hili kama ifuatavyo:anzaalignfrac12 ge frac12, frac13 frac14 frac14 frac14 frac12, frac15 frac16 frac17 frac18 frac18 frac1 8 frac18 frac18 frac12,mwishoalignna kadhalika. Hivyo,1 frac12 frac13 frac14 dotsb 1 frac12 frac1 2 dotsb,ambayo inaonyesha kuwa H_n kwa infty kama n to infty. 