Theorem 2 − 17 : Let { a n } be a bounded sequence of real numbers. Then: (a) lim sup a n = L if and only if, for any ε &gt; 0 , there are infinitely many terms of { a n } in ( L − ε , L + ε ) but only finitely many terms of { a n } with a n &gt; L + ε . Proves directly that lim sup { a n + b n } ≤ lim sup { a n } + lim sup { b n }

Question

Theorem           2          −          17          :           Let                       {                          a                              n                                      }                     be a bounded sequence of real numbers. Then:                                                                                                            (a)                   lim sup                                                            a                                              n                                                                              =                  L                   if and only if, for any                   ε                  &amp;gt;                  0                  ,                   there are infinitely many terms of                                                                                                                           {                                          a                                              n                                                              }                                     in                   (                  L                  −                  ε                  ,                  L                  +                  ε                  )                   but only finitely many terms of                                       {                                          a                                              n                                                              }                                     with                                       a                                          n                                                        &amp;gt;                  L                  +                  ε                  .                                                                        Proves directly that   lim sup  {      a    n    +      b    n    }  ≤  lim sup  {      a    n    }  +  lim sup  {      b    n    }

Payton Mcbride · Accepted Answer

Suppose that exists   c  &amp;gt;  0 such that      lim sup    n        a    n    +      b    n    =      lim sup    n        a    n    +      lim sup    n        b    n    +  ccall   A  :=      lim sup    n        a    n    ,        B  :=      lim sup    n        b    n  , then by the theorem exists       n    0   such that  ∀  n  &amp;gt;      n    0    :         b    n    ≤  Bso  ∀  n  &amp;gt;      n    0    :         a    n    +      b    n    ≤  B  +      a    n  then,  c  =      lim sup    n        a    n    +      b    n    −  B  −  A  =      lim sup          n      &amp;gt;              n        0                  a    n    +      b    n    −  A  −  B  ≤      lim sup    n        a    n    +  B  −  B  −  A  =contradiction

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