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Cauchy Product of Summable Sequences is Cesàro Summable

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According to Wikipedia, if $a_\bullet,b_\bullet$ are real (or complex) sequences such that $\sum_n a_n\to A$ and $\sum_n b_n\to B$, then$$\boxed{\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N\sum_{i=1}^n\sum_{k=0}^i a_{k}b_{i-k}=AB}$$Do you know where I can find a proof of this result? If not, I would also be satisfied with a self-contained proof of the following immediate corollary which I haven't been able to prove

Corollary: If $\sum_na_n\to A,\sum_nb_n\to B, \sum_na'_n\to A', \sum_nb'_n\to B'$, then it follows that$$\boxed{\left(\sum_{k=0}^n a_kb_{n-k}\right)_{n\geq0}=\left(\sum_{k=0}^na'_kb'_{n-k}\right)_{n\geq0}\Rightarrow AB=A'B'}$$

Thanks in advance! (This is not homework, btw. I was just on that Wikipedia page and that statement piqued my curiosity.)


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