Prove that if a n → L then the sequence b n = 1 / n 2 ⋅ ( a 1 + 2 a 2 + 3 a 3 + . . . + n a n ) converges to 0.5L

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Prove that if        a    n    →  L then the sequence       b    n    =  1      /        n    2    ⋅  (      a    1    +  2      a    2    +  3      a    3    +  .  .  .  +  n      a    n    ) converges to 0.5L

Madisyn Avery · Accepted Answer

Let   ϵ  &amp;gt;  0 be arbitrary, then find   N  ∈      N   so that   n  ≥  N implies   L  −  2  ϵ  &amp;lt;      a    n    &amp;lt;  L  +  2  ϵ for   n  ≥  NThis implies, for    n  ≥  N , that      ∑          k      =      1              N      −      1        k      a    k    +      ∑          k      =      N        n    k  (  L  −  2  ϵ  )  &amp;lt;      ∑          k      =      1        n    k      a    k    &amp;lt;      ∑          k      =      1              N      −      1        k      a    k    +      ∑          k      =      N        n    k  (  L  +  2  ϵ  )We can write the above inequalities as            [            ∑          k      =      1              N      −      1        k      a    k    −      ∑          k      =      1              N      −      1        k  (  L  −  2  ϵ  )            ]        +      ∑          k      =      1        n    k  (  L  −  2  ϵ  )  &amp;lt;      ∑          k      =      1        n    k      a    k    &amp;lt;            [            ∑          k      =      1              N      −      1        k      a    k    −      ∑          k      =      1              N      −      1        k  (  L  +  2  ϵ  )            ]        +      ∑          k      =      1        n    k  (  L  +  2  ϵ  )Divide by       n    2        1          n      2                  [            ∑          k      =      1              N      −      1        k      a    k    −      ∑          k      =      1              N      −      1        k  (  L  −  2  ϵ  )            ]        +      ∑          k      =      1        n        1    n    ⋅      k    n    (  L  −  2  ϵ  )  &amp;lt;      1          n      2            ∑          k      =      1        n    k      a    k    &amp;lt;      1          n      2                  [            ∑          k      =      1              N      −      1        k      a    k    −      ∑          k      =      1              N      −      1        k  (  L  +  2  ϵ  )            ]        +      ∑          k      =      1        n        1    n    ⋅      k    n    (  L  +  2  ϵ  )Take   n  ⟶  ∞ and use squeeze theorem to deduce      ∫    0    1    (  L  −  2  ϵ  )  x      d    x  ≤      lim          n      →      ∞            1          n      2            ∑          k      =      1        n    k      a    k    ≤      ∫    0    1    (  L  +  2  ϵ  )  x      d    xThis is equivalent to  0.5  L  −  ϵ  ≤      lim          n      →      ∞            1          n      2            ∑          k      =      1        n    k      a    k    ≤  0.5  L  +  ϵThe result follows since   ϵ  &amp;gt;  0 is arbitrary.

Prove that if a n </msub> → L then the sequence b n

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