If a n → a then b n = 1 n 2 ∑ i = 1 n i a i → a / 2

Question

If       a    n    →  a then       b    n    =      1          n      2            ∑          i      =      1        n    i      a    i    →  a      /    2

glapaso7ng5 · Accepted Answer

Let   ϵ  &amp;gt;  0 be given. By definition, there exists   N  ∈      N   such that   n  ≥  N implies       |        a    n    −  a      |    &amp;lt;  ϵ. This then implies   a  −  ϵ  &amp;lt;  a  &amp;lt;  a  +  ϵ and we have      ∑          i      =      1        n    i      a    i    =      ∑          i      =      1        N    i      a    i    +      ∑          i      =      N      +      1        n    i      a    i    ⇒      ∑          i      =      N      +      1        n    i  (  a  −  ϵ  )  &amp;lt;      ∑          i      =      1        n    i      a    i    −      ∑          i      =      1        N    i      a    i    &amp;lt;      ∑          i      =      N      +      1        n    i  (  a  +  ϵ  )  ⇒  (  a  −  ϵ  )      1    2    (  n  −  N  )  (  n  +  N  +  1  )  &amp;lt;      ∑          i      =      1        n    i      a    i    −      ∑          i      =      1        N    i      a    i    &amp;lt;  (  a  +  ϵ  )      1    2    (  n  −  N  )  (  n  +  N  +  1  )This then implies  (  a  −  ϵ  )            (      n      −      N      )      (      n      +      N      +      1      )              2              n        2              &amp;lt;      1          n      2            ∑          i      =      1        n    i      a    i    −      1          n      2            ∑          i      =      1        N    i      a    i    &amp;lt;  (  a  +  ϵ  )            (      n      −      N      )      (      n      +      N      +      1      )              2              n        2            Taking limits as n approaches infinity gives            a      −      ϵ        2    =      lim          n      →      ∞        (  a  −  ϵ  )            (      n      −      N      )      (      n      +      N      +      1      )              2              n        2              &amp;lt;      lim          n      →      ∞            b    n    +      [          1              n        2                    ∑              i        =        1            N        ]    i      a    i        lim          n      →      ∞            1          n      2      Since the limit of       1          n      2       is 0 this simplifies to            a      −      ϵ        2    &amp;lt;      lim          n      →      ∞            b    n    &amp;lt;            a      +      ϵ        2  Since   ϵ was arbitrary we conclude      a    2    ≤      lim          n      →      ∞            b    n    ≤      a    2  We conclude      lim          n      →      ∞            b    n    =      a    2

If a n </msub> → a then b n </msub> = 1

Answered question

Answer & Explanation

New Questions in Calculus 2