调和级数

hbghlyj

$\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} {i^2}\left( {1 + \frac{1}{2} + \frac{1}{3} +  \cdots  + \frac{1}{i}} \right) = \frac{{n\left( {n + 1} \right)}}{{36}}\left( { - 4n - 5 + 6\left( {2n + 1} \right)\left( {1 + \frac{1}{2} + \frac{1}{3} +  \cdots  + \frac{1}{{m + 1}}} \right)} \right)$
$\mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} {i^3}\left( {1 + \frac{1}{2} + \frac{1}{3} +  \cdots  + \frac{1}{i}} \right) =\frac{{n\left( {n + 1} \right)}}{{48}}\left( { - \left( {n + 2} \right)\left( {3n + 1} \right) + 12n\left( {n + 1} \right)\left( {1 + \frac{1}{2} + \frac{1}{3} +  \cdots  + \frac{1}{{n + 1}}} \right)} \right)$

tommywong

我有公式㗎啵 睇吓吖

artofproblemsolving.com/community/c728438h170 … tion_with_polynomial

$\displaystyle\sum_{k=1}^n H_k p(k)=(\sum_{m=0}^{deg(p)} \binom{n+1}{m+1} \Delta^m p(0))H_n
-\sum_{m=0}^{deg(p)}\frac{1}{m+1}\binom{n}{m+1}\Delta^m p(0)$

$p(k)=k^2,~\Delta p(k)=2k+1,~\Delta^2 p(k)=2$

$\displaystyle\sum_{k=1}^n H_k k^2=\left(\binom{n+1}{2}+2\binom{n+1}{3}\right)H_n
-\left(\frac{1}{2}\binom{n}{2}+\frac{2}{3}\binom{n}{3}\right)$

$p(k)=k^3,~\Delta p(k)=3k^2+3k+1,~\Delta^2 p(k)=6k+6,~\Delta^3 p(k)=6$

$\displaystyle\sum_{k=1}^n H_k k^3=\left(\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}\right)H_n
-\left(\frac{1}{2}\binom{n}{2}+\frac{6}{3}\binom{n}{3}+\frac{6}{4}\binom{n}{4}\right)$

hbghlyj

证明：存在正常数C,使得对任意正整数m,任意m个正整数$a_1,a_2,\cdots,a_m$,都有$H(a_1)+H(a_2)+\cdots+H(a_n)\leq C\sqrt{\sum\limits_{i=1}^m ia_i},H(n)=1+\frac12+\cdots+\frac1n$

hbghlyj

$\prod_{k=1}^n\left(\frac{\sum_{p=1}^k\frac1{2p-1}}{\sum_{p=1}^k\frac1{p}}\right)\ge \frac{n+1}{2^n}$
证明：只需证局部不等式$\frac{\sum_{p=1}^k\frac1{2p-1}}{\sum_{p=1}^k\frac1{p}}\ge\frac{k+1}{2k}\Leftrightarrow\frac {H(2n)-\frac12H(n)}{H(n)}>\frac{2n+1}{2n}\Leftrightarrow\frac {H(2n)}{H(n)}>\frac{2n+1}{2n}\Leftrightarrow(2n+1)H(2n)\ge n\ge 2nH(n)$,当且仅当n=1时取等