|
$\quad\boxed{ e = 3^{(1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\cdots)^{-1}}=4^{(1+\frac{1}{2}+\frac{1}{3}-\frac{3}{4}+\cdots)^{-1}}=\cdots}$dc ©Elinkage数学论坛 -- Elinkage极酷超级论坛 #6 题:对$\small\,\mathbb{N}\ni k>1,\;\,$令$\;a_n={\small\begin{cases}\frac{1}{n},& k\not\mid n,\\ -\frac{k-1}{n}, & k\mid n. \end{cases}}\quad$则$\;\;\boxed{e = k^{(a_1+a_2+a_3+\cdots)^{-1}}}$b= 证:记$\;m = \lfloor\frac{n}{k}\rfloor,\,$则$\;S_n = a_1+\cdots +a_n=\small\displaystyle\sum_{u=1}^n\frac{1}{u}-\sum_{u=1}^m\frac{1}{u}=H_n -H_m$}B({nJ $\underset{\,}{\because}\quad H_n = \log n+\gamma+O(\frac{1}{n})\quad$(ref)[4L $\underset{\,}{\therefore}\quad S_n = \ln\frac{n}{m}+O(\frac{1}{n}),\quad\displaystyle\lim_{n\to\infty}S_n = \ln k$l:Qs\ $\therefore\quad k^{(a_1+a_2+\cdots)^{-1}}= k^{(\ln k)^{-1}}=e^{(\ln k)(\ln k)^{-1}}=e.\quad\square$b-
| | |
|
|