Prove that if a n </msub> &#x2192;<!-- → --> L then the sequence b n

Bailee Landry 2022-05-24 Answered
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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Answers (1)

Madisyn Avery
Answered 2022-05-25 Author has 12 answers
Let ϵ > 0 be arbitrary, then find N N so that n N implies L 2 ϵ < a n < L + 2 ϵ for n N
This implies, for n N , that
k = 1 N 1 k a k + k = N n k ( L 2 ϵ ) < k = 1 n k a k < 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 ) < k = 1 n k a k < [ 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 ) < 1 n 2 k = 1 n k a k < 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 x
This is equivalent to
0.5 L ϵ lim n 1 n 2 k = 1 n k a k 0.5 L + ϵ
The result follows since ϵ > 0 is arbitrary.
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