HMMT 二月 2010 · CALC 赛 · 第 10 题

英文解析

10. [ 8 ] Let f ( n ) = . Then there exists constants γ , c , and d such thatkk =1
    c d 1
    f ( n ) = ln( n ) + γ + + + O ( ) ,
    2 3
    n n n
    1 1
    where the O ( ) means terms of order or lower. Compute the ordered pair ( c, d ).
    3 3
    n n
    1 1 k 1
    Answer: ( , − ) From the given formula, we pull out the term from O ( ), making f ( n ) =
    3 4
    2 12 n nc d k 1
    log( n ) + γ + + + + O ( ). Therefore,
    2 3 4
    n n n n
    ( ) ( ) ( ) ( ) ( )
    n + 1 1 1 1 1 1 1 1
    f ( n +1) − f ( n ) = log − c − − d − − k − + O .
    2 2 3 3 4
    n n n + 1 n ( n + 1) n ( n + 1) n
    1 1
    For the left hand side, f ( n + 1) − f ( n ) = . By substituting x = , the formula above becomesn +1 nx 1 x + 2 x + 3 x + 32
    2 3 4 4 = log (1 + x ) − cx · − dx · − kx · + O ( x ) .
    2 3
    x + 1 x + 1 ( x + 1) ( x + 1)
    1 1
    Because x is on the order of , is on the order of a constant. Therefore, all the terms in then ( x +1)3
    x +3 x +32
    4 4 4
    expansion of kx · are of order x or higher, so we can collapse it into O ( x ). Using the Taylor
    ( x +1)3
    expansions, we get
    ( )
    ( ) ( ) ( )
    1 1
    2 4 2 3 2 3 4
    x 1 − x + x + O x = x − x + x − cx (1 − x ) − dx (2) + O x .
    2 3
    1 1
    Coefficient comparison gives c = and d = − .
    2 12
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