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If $a_n \to a$ and $b_n \to b$, then $\sum_{k=1}^n \frac{a_kb_{n-k}}{n} \to ab$.

Is this an application of Cesàro series? If the term $\sum_{k=1}^n {a_kb_{n-k}}$ is a partial sum of some sequence that converges to $ab$. How to do this?

metamorphy
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jerry guna
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  • Sketch: Pick $\epsilon>0$, find $m_0, n_0$ big enough so that $|a_mb_n-ab|<\epsilon/2$ for any $m\ge m_0, n\ge n_0$, then take $N\ge m_0+n_0$ and estimate $\sum_{k=1}^{N}\frac{a_k b{n-k}}{N}$. That sum will have $m_0-1$ terms where $k<m_0$ and $n_0$ terms where $n-k<n_0$ and all the other terms will be ($\epsilon/2$) "close" to $ab$, so by now picking $N$ "big enough" the whole sum should be up to $\epsilon$ away from $ab$. –  Sep 16 '21 at 10:23
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    WLOG, $a=b=0$, then $$\left| \sum_{k=1}^n \frac{a_kb_{n-k}}{n} \right| \le \sqrt{ \frac{1}{n} \sum_{k=1}^n a_k^2} \sqrt{ \frac{1}{n} \sum_{k=1}^n b_k^2} \rightarrow 0 $$ – Paresseux Nguyen Sep 16 '21 at 10:26
  • (by the way, I guess the sum should either go from $0$ to $n$ or from $1$ to $n-1$, otherwise the implication is that $a_n$ starts from $a_1$ but $b_n$ starts from $b_0$. –  Sep 16 '21 at 10:27

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