So, I have this problem lim n → ∞ 1 n + ( − 1 ) n n 2 = 0

Question

So, I have this problem      lim          n      →      ∞            1    n    +            (      −      1              )        n                    n      2        =  0

Ada Harrington · Accepted Answer

We want to show that                    (1)                              lim                      n            →            ∞                          (                  1          n                +                              (            −            1                          )              n                                            n            2                          )        ,            so we have to show that for each   ε  &amp;gt;  0 there exists an       N    ε   such that                    (2)                              |                      1            n                    +                                    (              −              1                              )                n                                                    n              2                                |                &amp;lt;        ε        ,                ∀        n        ∈                  N                :                n        &amp;gt;                  N          ε                    We use N with the subscript ε to make clear that N depends on ε. Different values of ε need different values of N.So given   ε, we want to calculate       N    ε   such that (2)  holds. But calculating tusch an       N    ε   directly from (2)  is difficult because the LHS (left hand side) of the inequality (2) is a complicated expression. But we can help us. We do not need the least possible       N    ε  . So we can substitute the LHS of (2)  by an expression that is greater than       |          1      n        +                  (        −        1                  )          n                            n        2              |  . If this greater expression is less than   ε then       |          1      n        +                  (        −        1                  )          n                            n        2              |   is less than this greater expression and therefore less than   εWe know that the following hold for real numbers   a  ,  b      |    a  +  b      |    ≤      |    a      |    +      |    b      |        |    a  b      |    =      |    a      |        |    b      |  and from the latter follows      |        a    n        |    =      |    a            |        k    ,  ∀  k  ∈      N  and      |          a      b        |    =                    |            a              |                            |            b              |            So we have                    (3)                              |                      1            n                    +                                    (              −              1                              )                n                                                    n              2                                |                ≤                  |                      1            n                    |                +                  |                                    (              −              1                              )                n                                                    n              2                                |                =                                            |                        1                          |                                                          |                        n                          |                                      +                                            |                        (            −            1            )                                          |                            n                                            |                          n              2                        |                                  =                  1                                    |                        n                          |                                      +                  1                                    |                                      n              2                                      |                                      =                  1          n                +                  1                      n            2                          ≤                  1          n                +                  1          n                =                  2          n                    The last   ≤ hold because   n  ≤      n    2   and so       1          n      2        ≤      1    n  . So instead of (2) we now try to solve                    (4)                              2          n                &amp;lt;        ε        ,                ∀        n        ∈                  N                :                n        &amp;gt;                  N          ε                    This is easy, we get                    (5)                    n        &amp;gt;                  2          ε                    So whenever (5) holds, (4) holds, too. And because of      |          1      n        +                  (        −        1                  )          n                            n        2              |    ≤      2    n    ,that follows by (3), (2) holds, too. So because of (5) we choose      N    ε    =      2    ε  and then (2) holds.

Carlie Fernandez · Accepted Answer

Notice that  (  −  1      )    n    =      {                            1          ,                          if           n           even                                      −          1          ,                          if           n           odd                        and therefore       |    (  −  1      )    n        |    =  1. A useful trick to solve your problem is given by the triangle inequality (you probably came across this when proving the uniqueness of limits in a Calculus course):       |    a  +  b      |    ≤      |    a      |    +      |    b      |  . Thus:                    (1)                              |                      1            n                    +                                    (              −              1                              )                n                                                    n              2                                |                ≤                  1          n                +                  1                      n            2                          .            You need to prove that the above quantity is smaller than   ϵ when n is large enough. So, you can use (1) to find   M  ∈      R   such that      |          1      n        +                  (        −        1                  )          n                            n        2              |    ≤  ϵ   when   n  ≥  M  .

So, I have this problem <munder> <mo movablelimits="true" form="prefix">lim <mrow class=

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New Questions in Calculus 1