# Limit of a sequence

#### Fernando Revilla

##### Well-known member
MHB Math Helper
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 $$L=ab$$.

#### Poly

##### Member
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

##### Well-known member
MHB Math Helper
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.