Question

18. If ai, az, as,... is a bounded sequence of real numbers, define lim sup an (also denoted lim an) to be --+ n+ l.u.b. {z ER: an > & for an infinite number of integers n} and define lim inf an (also denoted lim an) to be g.l.b. {ER: An <for an infinite number of integers n}. Prove that lim inf an Slim sup an, with the equality holding if and only if the sequence converges. 19. Let ai, as, as, ... and bı, bz, ba, ... be bounded sequences of real numbers. Show that lim sup (an + bn) <lim sup An + lim sup bn, with the equality holding if one of the original sequences converges (cf. Prob. 18).

Answer 1

13 pase to an tbn < atbt e so <antbr> <arbre areatest limit point of contbon) = ath - - lim <ant bn> < atb lim cant bn> s limcast lim <br>

1st page given that a, az, .... is a bounded sequence of seal numbers let x = {MER: an>x Lor infinitely many na Lim sup ana limane lub{x} according to definition of lub and let l be lub of then is for every given Exo an>l-E. for infinitely ronny no and 1) Lor every & so an alte eventually so our conclusion that dis lim an iss it satisfied both conditions Simillarly dis lim an if (1) for every exo anal te for infinitely many n {(ii) for every Exo an>d-e except finitely many n (eventually) we also conclude that ' is limit boint but but no other limit boint less than i .. because an has finitely many value Such that an 1-2

1 2nd pase From 2nd condition of definition of lim an and lin an 1-{<an & An< lte except finitely many value of a Here & f ' are lim and lim. of any eventually 1-4. < an alte and & → l's 1 proved let lim Sup An= a tlim sup bu= b. le for every & so there exist m, and me Such that both condition of lim (and holds le On <a+& nom. And by <b+& nom we take max {m, m} = m (day) then both an eat & , bn <bt & rnem.
on first page in definition of Lim inf <an> int both condition write l' in place of l.