每日一题[958]裂项求和与放缩

求证:$\displaystyle \sum_{k=1}^{n}\dfrac{16}{(2k+1)(2k+2)}>\dfrac{9n-3}{4n+3}$.


cover分析与解 裂项放缩,有\[LHS=\sum_{k=1}^{n}\dfrac{16}{(2k+1)(2k+2)}>\sum_{k=1}^{n}\dfrac{16}{(2k+1)(2k+3)}=\dfrac 83-\dfrac 8{2n+3}>\dfrac{9n-3}{4n+3}.\]事实上,当$n\geqslant 4$时,有\[LHS\geqslant \dfrac{16}{12}+\dfrac{16}{30}+\dfrac{16}{56}+\dfrac{16}{90}=\dfrac{734}{315}>\dfrac 94>RHS.\]

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每日一题[958]裂项求和与放缩》有 2 条评论

  1. thebluesky thebluesky说:

    $$16\sum_{k=1}^n\left(\dfrac 1{2k+1}-\dfrac 1{2k+2}\right)和16\left(\dfrac 13-\dfrac 1{2n+2}\right)不相等吧,$$
    $$应该是\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+…+\frac{1}{2n+1}-\frac{1}{2n+2}吧?$$

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