Suppose that $a_n$ and $b_n$ converge to $\alpha$ and $\beta$ as $n\to\infty$ respectively. Show that: $$\frac{a_0b_n+a_1b_{n-1}+…+a_nb_0}{n}$$
converges to $\alpha\beta$ as $n\to\infty$
Resolution: Let $M$ be an upper bound of the two convergent sequences $|a_n|$ and $|b-n|$. For any $\epsilon>0$ we can take a positive integer $N$ satisfying $|a_n-\alpha|<\epsilon$ and $|b_n-\beta|<\epsilon$ for all integers $n>N$. If $n>N^2$, then:
$|a_k-b_{n-k}-\alpha\beta|\leqslant|(a_k-\alpha)b_{n-k}+\alpha(b_{n-k}-\beta)|\leqslant(M+|\alpha|)\epsilon$
for any integer $k\in[\sqrt{n},n-\sqrt{n}]$. Therefore:
$$|\frac{1}{n}\sum_\limits{k=0}^{n}a_kb_{n-k}-\alpha\beta|\leqslant\frac{1}{n}\sum_\limits{\sqrt{n}\leqslant k\leqslant n-\sqrt{n}}^{n}a_kb_{n-k}-\alpha\beta\\+2(\alpha\beta+M^2)\frac{\sqrt{n}+1}{n}\leqslant(M+|\alpha|)\epsilon+2(\alpha\beta+M^2)$$
we can take $n$ so large that the last expression is less than $(M+|\alpha|+1\epsilon$
Questions:
1)
I have just copied the resolution. However I think that here $\frac{1}{n}\sum_\limits{\sqrt{n}\leqslant k\leqslant n-\sqrt{n}}^{n}a_kb_{n-k}-\alpha\beta\\+2(\alpha\beta+M^2)\frac{\sqrt{n}+1}{n}$
there should not be a sum once the author is using$2(\alpha\beta+M^2)\frac{\sqrt{n}+1}{n}$. What do you think?
2) How did the author derive $\frac{\sqrt{n}+1}{n}$?
Best Answer
The author divides the sum $\sum_{k=0}^n$ into two: one with $k\notin[\sqrt{n},n-\sqrt{n}]$ and one with $k\in[\sqrt{n},n-\sqrt{n}]$. The first one has at most $2\,(\sqrt{n}+1)$ terms (from $0$ to $\sqrt n$ and from $n-\sqrt n$ to $n$), and each term is bounded by $M^2+|\alpha\,\beta|$. Thus$$ \Bigl|\frac{1}{n}\sum_{k\notin[\sqrt{n},n-\sqrt{n}]}a_kb_{n-k}-\alpha\beta\,\Bigr|\le\frac{1}{n}\,2\,(\sqrt{n}+1)(M^2+|\alpha\,\beta|). $$ For the second each term is a most (M+|\alpha|)\,\epsilon and there are less than $n$ terms. Then $$ \Bigl|\frac{1}{n}\sum_{k\in[\sqrt{n},n-\sqrt{n}]}a_kb_{n-k}-\alpha\beta\,\Bigr|\le\frac{1}{n}\,\Bigl(\sum_{k\in[\sqrt{n},n-\sqrt{n}]}1\Bigr)(M+|\alpha|)\,\epsilon\le(M+|\alpha|)\,\epsilon. $$