Question 8.1: (a) It’s obvious that u n and therefore U  lim un is either finite or  . n  If U is finite, then for   0 , there exists N  such that U  un  U   for n  N . Hence, un  U   implies that an  U   for n  N . Given    0 and m  0 , there exists n  m such that U     an by U  un  sup ak : k  n .Hence, we obtain U  lim sup an . n  If U   , then for any , there exists a positive integer N such that as n ≥ N, we have un   M , which implies that an ≤ −M for n ≥ N. Hence, lim an   That is, an  is not n  bounded below. If an  has a finite limit supremum a, then for   0 and m  0 , there exists n > m such that an  a   , which contradicts to lim an   . Hence, U  lim sup an . n  n  (b) Similarly (c) Since U  lim sup an , we have n  For   0 , there exists N  such that an  U   as n  N For   0 and m  0 , there exists an integer P  m  m such that U    aP m   Hence, aP m  is a convergent subsequence of an  that converges to U. Similarly for the case of V. (d) For   0 , there exists a positive integer N1 such that as n ≥ N1, we have an  U   , and there exists N 2  such that as n  N 2 , we have U    an . Therefore, for n  max  N1 , N2  , we have U  ...
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