Let $|q|<1$, and $(t_i)_{i\geq 0}$ a sequence converging to 0. Why is $\lim_{n\to \infty}\sum_{i=0}^{n} q^i \cdot t_{n-i}=0$?
Convergence of series based on convergence of sequence
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0Oops, misread one important piece of info. Never mind this. – 2012-05-20
2 Answers
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
$\begin{align*}\left|\sum_{k=0}^n q^kt_{n-k}\right|&<\frac{\epsilon}2+|q|^{n-n_0+1}\sum_{k=0}^{n_0-1}|t_k|\\ &<\frac{\epsilon}2+\frac{\epsilon}{2Mn_0}\sum_{k=0}^{n_0-1}M\\ &\le\epsilon\;. \end{align*}$
It follows that $\displaystyle\lim_{n\to\infty}\sum_{k=0}^nq^kt_{n-k}=0$.
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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1$C_1=\sup_K \left| \sum_{i=0}^K q_i\right|$, $C_2=\sup_i |t_i|$. These are finite since $(t_i)$ and \sum_{i=0}&\infty q_i are convergent. They are fixed numbers and do not depend on $n$. – 2012-05-20