求证：无穷连乘积 ∏(k=3,∞)k(3k-1)／[(k-1)(3k+2)]=4／3

王守恩

求证：\(\displaystyle\prod_{k=3}^{\infty}\frac{k(3k-1)}{(k-1)(3k+2)}=\frac{4}{3}\)

Future_maths

\(=\lim_{n\to+\infty}\left\{ \frac{3\times8}{2\times11}\cdot\frac{4\times11}{3\times14}\cdot\frac{5\times14}{4\times17}\cdot\frac{6\times17}{5\times20}\cdot\cdot\cdot\frac{n\left( 3n-1\right)}{\left( n-1\right)\left( 3n+2\right)}\cdot\frac{\left( n+1\right)\times\left( 3n+2\right)}{n\times\left( 3n+5\right)}\cdot\frac{\left( n+2\right)\times\left( 3n+5\right)}{\left( n+1\right)\times\left( 3n+8\right)}\right\}\)

Future_maths

\(=\lim_{n\to+\infty}\left\{ \frac{8}{2}\cdot\frac{n+2}{3n+8}\right\}=\frac{4}{3}\)

王守恩

Future_maths 发表于 2022-11-8 16:56
\(=\lim_{n\to+\infty}\left\{ \frac{3\times8}{2\times11}\cdot\frac{4\times11}{3\times14}\cdot\frac{5\ ...

谢谢 Future_maths 点拨，出题的就是搞复杂也不怕了。

求证：\(\displaystyle\prod_{k=a}^{\infty}\frac{k(2ak-k-a)}{(k-1)(2ak-k+a-1)}=\frac{2a}{2a-1}\)

时空伴随者

\(\displaystyle\prod_{k=3}^{\infty}\frac{k(3k-1)}{(k-1)(3k+2)}\)
\(=\displaystyle\prod_{k=3}^{\infty}\frac{3+\frac2{k-1}}{3+\frac2k}\)
\(=(3+\frac2{3-1})\displaystyle\lim_{k \to \infty}\frac1{3+\frac2k}\)
\(=\frac43\)

时空伴随者

\(\displaystyle\prod_{k=a}^{\infty}\frac{k(2ak-k-a)}{(k-1)(2ak-k+a-1)}\)
\(=\displaystyle\prod_{k=a}^{\infty}\frac{2a-1+\frac{a-1}{k-1}}{2a-1+\frac{a-1}k}\)
\(=(2a-1+\frac{a-1}{a-1})\displaystyle\lim_{k \to \infty}\frac1{2a-1+\frac{a-1}k}\)
\(=\frac{2a}{2a-1}\)

luyuanhong

楼上 时空伴随者 的解答很好！已收藏。