2023.9.29 闲话

APJ 让我鲜花被投我要魔怔杯了 😦

Her opinion on the disaster destroying all is delivered in a single word.

「Furry」

😭回来吧 BOBO😭
🌟我最骄傲的信仰🌟
⚡️历历在目的演讲⚡️
😭眼泪莫名在流淌😭
💥依稀记得模拟赛💥
👍还有给力的学长👍
⚡️把贡品都给打退⚡️
✨通宵熬夜都不累✨


证明:

\[\sum_{k=1}^n\dfrac{(-1)^{k+1}}k\binom{n}{k}=H_n \]

(具体数学 6.72)

纯净做法

注意到:

\[\frac{1}{k}\binom{n}{k}=\frac{1}{k}\left(\binom{n-1}{k}+\binom{n-1}{k-1}\right) =\frac{1}{k}\binom{n-1}{k}+\frac{1}{n}\binom{n}{k} \]

\(\displaystyle F(n)=\sum_{k=1}^n\dfrac{(-1)^{k+1}}k\binom{n}{k}\),那么可以展开:

\[\begin{aligned}F(n)&=\sum_{k=1}^{n-1}\dfrac{(-1)^{k+1}}k\dbinom nk+\dfrac{(-1)^{n+1}}n\\&=\sum_{k=1}^{n-1}\dfrac{(-1)^{k+1}}k\dbinom{n-1}k+\dfrac1n\sum_{k=1}^{n-1}(-1)^{k+1}\dbinom nk+\dfrac{(-1)^{n+1}}n\\&=F(n-1)+\dfrac1n\sum_{k=1}^n(-1)^{k+1}\dbinom nk\\&=F(n-1)+\dfrac1n\end{aligned} \]

从而 \(F(n)=H_n\) .

你先别急

左边的 OGF 有平凡的刻画:

\[F(z)=\int\dfrac{1-(1-z)^n}z\mathrm dz \]

所以:

\[\begin{aligned}\mathrm{LHS}&=\int_0^1\dfrac{1-(1-z)^n}z\mathrm dz\\&=\int_0^1(1-(1-z)^n)\mathrm d(\ln z)\\&=-n\int_0^1\ln z(1-z)^{n-1}\mathrm dz\\&=-n\int_0^1z^{n-1}\ln(1-z)\mathrm dz\\&=n\int_0^1z^{n-1}\sum_{k\ge1}\dfrac{z^k}k\mathrm dz\\&=\sum_{k\ge1}\dfrac nk\int_0^1z^{n+k+1}\mathrm dz\\&=\sum_{k\ge1}\dfrac n{k(n+k)}\\&=H_n=\mathrm{RHS}\end{aligned} \]

And, The Deliverance of Light.

posted @ 2023-09-29 06:16  Jijidawang  阅读(181)  评论(21编辑  收藏  举报
😅​