MHB Proving the Limit of a Sequence with Toeplitz's Theorem

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The discussion focuses on proving the limit of a specific sequence using Toeplitz's theorem, but participants express that it may not be applicable. The limit to prove is L = lim(n→∞) (a1b_n + a2b_{n-1} + ... + a_nb_1)/n = ab, given that lim(n→∞) a_n = a and lim(n→∞) b_n = b. The proof involves manipulating the sum and applying the Arithmetic Mean Criterion to establish the relationship between the terms. Participants seek clarification on specific steps in the proof, particularly regarding the application of inequalities and limit properties. The conversation emphasizes the need for a deeper understanding of the limit proof process.
Fernando Revilla
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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.
 
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Fernando Revilla said:
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?
 
Poly said:
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.
 
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