求解无穷乘积问题一例
`\D\prod^\infty_{k=2}\left(1-\frac1{\sum^k_{t=1}t}\right)` $$\prod^\infty_{k=2}\left(1-\frac1{\sum^k_{t=1}t}\right)=\lim_{n\to \infty}\prod^n_{k=2}\frac{(k-1)(k+2)}{k(k+1)}=\lim_{n\to \infty}\frac{n+2}{3n}=\frac{1}{3}$$ 令`\D a_n=\frac{n+2}n`, 则`\D\prod^n_{k=2}\frac{(k-1)(k+2)}{k(k+1)}=\prod^n_{k=2}\frac{a_k}{a_{k-1}}=\frac{a_n}{a_1}=\frac{n+2}{3n}`
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