Fernando Revilla
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I quote an unsolved problem posted on December 9th, 2012 in another forum
Suppose \displaystyle\lim_{n \to \infty}a_n=a, \displaystyle\lim_{n \to \infty}b_n=b and without loss of generality that the sequences are (a_n)_{n\geq 0} and (b_n)_{n\geq 0}. We have to prove $L=\displaystyle\lim_{n\to \infty}\displaystyle\sum_{k=0}^n \frac {a_k b_{n-k}}{n}=ab$. We verify
\displaystyle\sum_{k=0}^n \frac {a_k b_{n-k} + ab}{n} = \sum_{k=0}^n \frac {(a_k - a)(b_{n-k} - b)}{n} +a \sum_{k=0}^n \frac {b_{n-k}}{n} + b \sum _{k=0}^n\frac {a_k}{n}
Taking limits and using the Arithmetic Mean Criterion we get L+ab=0+ab+ab, so L=ab.
Could someone help me prove the following?
\displaystyle\lim_{n \to \infty}\dfrac{a_1b_n+a_2b_{n-1}+\ldots+a_nb_1}{n}=ab
What theorem should I use. Toeplitz's theorem doesn't seem to be helpful.
Suppose \displaystyle\lim_{n \to \infty}a_n=a, \displaystyle\lim_{n \to \infty}b_n=b and without loss of generality that the sequences are (a_n)_{n\geq 0} and (b_n)_{n\geq 0}. We have to prove $L=\displaystyle\lim_{n\to \infty}\displaystyle\sum_{k=0}^n \frac {a_k b_{n-k}}{n}=ab$. We verify
\displaystyle\sum_{k=0}^n \frac {a_k b_{n-k} + ab}{n} = \sum_{k=0}^n \frac {(a_k - a)(b_{n-k} - b)}{n} +a \sum_{k=0}^n \frac {b_{n-k}}{n} + b \sum _{k=0}^n\frac {a_k}{n}
Taking limits and using the Arithmetic Mean Criterion we get L+ab=0+ab+ab, so L=ab.