回復 1# judy75 的帖子
一般項 \(\displaystyle\frac{1}{a_k a_{k+2}}=\left(\frac{1}{a_k}-\frac{1}{a_{k+2}}\right)\frac{1}{a_{k+2}-a_k}\)
\(\displaystyle=\left(\frac{1}{a_k}-\frac{1}{a_{k+2}}\right)\frac{1}{a_{k+1}}\)
\(\displaystyle=\frac{1}{a_k a_{k+1}}-\frac{1}{a_{k+1}a_{k+2}}\)
其中 \(k=1,2,3,\cdots\)
因此，\(\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{a_k a_{k+2}}=\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{1}{a_k a_{k+1}}-\frac{1}{a_{k+1}a_{k+2}}\right)\)
\(\displaystyle=\lim_{n\to\infty}\left(\frac{1}{a_1 a_2}-\frac{1}{a_{n+1}a_{n+2}}\right)\)
易知 \(\displaystyle\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}a_{n+2}=\infty\)［可證 \(a_k\geq n, \forall n\geq5\)］
因此，所求＝\(\displaystyle\frac{1}{a_1 a_2}=1\)
