Convergence of series based on convergence of sequence Let $|q|<1$, and $(t_i)_{i\geq 0}$ a sequence converging to 0. Why is $\lim_{n\to \infty}\sum_{i=0}^{n} q^i \cdot t_{n-i}=0$?
 A: 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.
A: 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 $\displaystyle\lim_{n\to\infty}\sum_{k=0}^nq^kt_{n-k}=0$.
