Let $|q|<1$, and $(t_i)_{i\geq 0}$ a sequence converging to 0. Why is $\lim_{n\to \infty}\sum_{i=0}^{n} q^i \cdot t_{n-i}=0$?
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Since $|q|<1$, we can let $\alpha=\sum_{n\ge 0}|q|^n$. Let $\epsilon>0$. Since $\langle t_n:n\in\Bbb N\rangle\to0$, there is an $n_0\in\Bbb N$ such that $|t_n|<\frac{\epsilon}{2\alpha}$ whenever $n\ge n_0$. Then for $n>n_0$ we have $$\begin{align*} \left|\sum_{k=0}^n q^kt_{n-k}\right|&\le\sum_{k=0}^n |q|^k|t_{n-k}|\\ &=\sum_{k=0}^{n-n_0}|q|^k|t_{n-k}|+\sum_{k=n-n_0+1}^n|q|^k|t_{n-k}|\\ &=\sum_{k=n_0}^n|q|^{n-k}|t_k|+\sum_{k=n-n_0+1}^n|q|^k|t_{n-k}|\\ &<\frac{\epsilon}{2\alpha}\sum_{k=n_0}^n|q|^{n-k}+|q|^{n-n_0+1}\sum_{k=0}^{n_0-1}|t_k|\\ &<\frac{\epsilon}2+|q|^{n-n_0+1}\sum_{k=0}^{n_0-1}|t_k|\;. \end{align*}$$ Note that the last summation is independent of $n$. Let $M=\max\{|t_k|:0\le k<n_0-1\}$. Since $|q|<1$, we can choose $m_0\in\Bbb N$ such that $|q|^{n-n_0+1}<\frac{\epsilon}{2Mn_0}$ whenever $n\ge m_0$. Then for $n>\max\{n_0,m_0\}$ we have $$\begin{align*}\left|\sum_{k=0}^n q^kt_{n-k}\right|&<\frac{\epsilon}2+|q|^{n-n_0+1}\sum_{k=0}^{n_0-1}|t_k|\\ &<\frac{\epsilon}2+\frac{\epsilon}{2Mn_0}\sum_{k=0}^{n_0-1}M\\ &\le\epsilon\;. \end{align*}$$ It follows that $\displaystyle\lim_{n\to\infty}\sum_{k=0}^nq^kt_{n-k}=0$. |
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Let us write $$ \sum_{i=0}^n q^i t_{n-i}=\sum_{i=0}^{K} q^i t_{n-i} + \sum_{i=K+1}^n q^i t_{n-i} $$ for some suitably chosen $K$, which will depend on $n$. For the first sum, note that $\sum_{i=0}^{K}q^i$ is bounded and that $t_{n-i}$ becomes small for all $i$ as long as $K$ is not too large (more precisely, if $n-K\to\infty$). For the second sum, we can use that $(t_i)$ is a bounded sequence and that $\sum_{i=K+1}^\infty q^i\to 0$ as $K\to\infty$. Taking $K=n/2$ should satisfy both conditions. |
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