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设 $lim_{n \to \infty} a_{n} = A$ , $lim_{n \to \infty} b_{n} = B$ , 则 $lim_{n \to \infty} \frac{1}{n} (a_{1} b_{n} + a_{2} b_{n - 1} + \dots + a_{n} b_{1}) = A B$ .

Posted by haifeng on 2014-05-14 22:48:09 last update 2014-05-14 22:48:09 | Edit | Answers (0)

设 $lim_{n \to \infty} a_{n} = A$ , $lim_{n \to \infty} b_{n} = B$ , 则
$lim_{n \to \infty} \frac{1}{n} (a_{1} b_{n} + a_{2} b_{n - 1} + \dots + a_{n} b_{1}) = A B .$

问题及解答

设 limn→∞an=A, limn→∞bn=B, 则 limn→∞1n(a1bn+a2bn−1+⋯+anb1)=AB.

设 $lim_{n \to \infty} a_{n} = A$ , $lim_{n \to \infty} b_{n} = B$ , 则 $lim_{n \to \infty} \frac{1}{n} (a_{1} b_{n} + a_{2} b_{n - 1} + \dots + a_{n} b_{1}) = A B$ .