Comment by poizan42

No, because the terms tends monotonically towards zero. Let an integer m with closest harmonic number H_n be given (i.e. n minimizes |H_n-m|). So m exists either between H_n and H_(n+1) or H_n and H_(n-1). Then |H_n-m| < H_(n+1) - H_(n-1) = 1/n + 1/(n+1). We can make that bound arbitrary small by choosing a large enough n.