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skylsn

skylsn

Answered question

2022-06-25

Given that ( a n ) n = 1 is a sequence for which a n / log ( n ) 1 as n , show that
lim x 1 n = 0 a n x n   1 1 x log ( 1 1 x ) = 1

Answer & Explanation

Donavan Mack

Donavan Mack

Beginner2022-06-26Added 24 answers

1 1 x log ( 1 1 x ) = n = 1 H n x n ,
where H n = 1 + 1 2 + + 1 n . Note that 0 < H n log n 1 holds. These are not so difficult to prove, so you should try doing so.
Combined with lim n a n log n = 1, we get lim n a n H n = 1 . So, for any ε > 0, there exists N so that n N implies | a n H n 1 | < ε
Fix N and let f ( x ) = n = 0 N 1 a n x n n = 1 N 1 H n x n for simplicity. Since N is a fixed finite number, note that f ( x ) is bounded near x = 1
| n = 0 a n x n 1 1 x log ( 1 1 x ) 1 | = | n = 0 a n x n n = 1 H n x n n = 1 H n x n | | f ( x ) | n = 1 H n x n + | n = N ( a n H n ) x n | n = 1 H n x n (by triangular inequality) | f ( x ) | n = 1 H n x n + n = N | a n H n | x n n = 1 H n x n (by triangular inequatlity) | f ( x ) | n = 1 H n x n + ε n = N H n x n n = 1 H n x n (by  | a n H n | < ε H n ) | f ( x ) | n = 1 H n x n + ε
Recalling that f(x) is bounded near x = 1 implies the RHS ε. Therefore we get
lim sup x 1 0 LHS ε
Since ε is arbitrary positive number, we conclude that
lim x 1 0 LHS = 0.

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