If an 0 and bn 0 prove that limsup an bn lim sup anlim sup

If a_n > 0 and b_n > 0, prove that limsup a_n b_n (lim sup a_n)(lim sup b_n).

Solution

Cannot solve witnout boundedness of an and bn

Solution: Suppose {an} and {bn} are bounded sequences of nonnegative numbers. For every m N, define Z _m := {a_nb_n : m > N}, X_m := {a_n : m > N}, and Ym := {b_n : m > N}. Since all the sequences are bounded, thus by the Completeness Axiom, their supremum must exist as real numbers. Let am = supZ_m , bm = supXm , and c_m= supY_m for every m N.

Now, Fix m N. If n m, thena_nb_n b_m c_m since a_n b_m , b_n c_m and a_n, b_n 0 n N,

so b_mc_m is an upper bound for Zm , and thus a_m b_m c_m . As m N is arbitrary, we have a_m b_m c_m m N. Taking the limit of both sides, we get lim m aN lim m b_mc_m = lim m b_m · lim m c_m . Therefore, we have lim sup n (a_nb_n) limn supa_n · limn supb_n