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Limit of a sequence

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I quote an unsolved problem posted on December 9th, 2012 in another forum

Could someone help me prove the following?

[tex]\displaystyle\lim_{n \to \infty}\dfrac{a_1b_n+a_2b_{n-1}+\ldots+a_nb_1}{n}=ab[/tex]

What theorem should I use. Toeplitz's theorem doesn't seem to be helpful.
Suppose [tex]\displaystyle\lim_{n \to \infty}a_n=a[/tex], [tex]\displaystyle\lim_{n \to \infty}b_n=b[/tex] and without loss of generality that the sequences are [tex](a_n)_{n\geq 0}[/tex] and [tex](b_n)_{n\geq 0}[/tex]. 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

[tex]\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}[/tex]

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

Poly

Member
Nov 26, 2012
32

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Could you elaborate on this step, please?
For the left side, [tex]\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[/tex]. For the first addend of the right side, use the inequality [tex]|xy|\leq \dfrac{x^2+y^2}{2}[/tex] and again the Arithmetic Mean Criterion.