I have to prove the following: inf ( 1 n + c o s ( π n ) | n ∈ N ) = − 1I thought that if I prove that inf ( A + B ) = inf ( A ) + inf ( B ) I can solve the question, since inf ( 1 n ) = 0 and inf ( π n ) = − 1, so the sum is −1Can I use this approach solving the question?

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I have to prove the following:  inf      (          1      n        +    c    o    s    (    π    n    )          |        n    ∈          N        )    =  −  1I thought that if I prove that   inf  (  A  +  B  )  =  inf  (  A  )  +  inf  (  B  ) I can solve the question, since   inf      (          1      n        )    =  0 and   inf  (  π  n  )  =  −  1, so the sum is −1Can I use this approach solving the question?

iberistazi · Accepted Answer

If you fix n odd, then you have the subsequence   −  1  +      1    n   and clearly   inf      (    −    1    +          1      n        )    =  −  1  +  inf      (          1      n        )  If you fix n even, then the subsequence is   1  +      1    n   which is always bigger than the other one so we don&#039;t really care about it.

I have to prove the following: <mo movablelimits="true" form="prefix">inf (

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