Proving the Limit of a Sequence with Toeplitz's Theorem

[Fernando Revilla]

I quote an unsolved problem posted on December 9th, 2012 in another forum

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.

 

[Poly1]

Taking limits and using the Arithmetic Mean Criterion we get L+ab=0+ab+ab, so L=ab.

Could you elaborate on this step, please?

 

[Fernando Revilla]

Could you elaborate on this step, please?

For the left side, \displaystyle\sum_{k=0}^n \frac {a_k b_{n-k} + ab}{n}=\displaystyle\sum_{k=0}^n \frac {a_k b_{n-k}}{n}+\dfrac{n+1}{n}ab\to L+ab. For the first addend of the right side, use the inequality |xy|\leq \dfrac{x^2+y^2}{2} and again the Arithmetic Mean Criterion.