An argument similar to the above (Exercise 30.8) shows that the subsequence (s_(2n+1)) is decreasing and bounded below by 0. Thus it also converges.Furthermore,since s_(2n+1)= s_2n+s_(2n+1) and lim〖a_(2n+1) 〗 = 0 , we must have lim〖s_(2n+1)= lim〖s_2n=s〗 〗