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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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Use the limit comparison test on $\sum_{i=0}^\infty q^i$ – Brett Frankel May 20 '12 at 16:44
Oops, I forgot to add the $\lim$. See edit. @Brett: like how? – akkkk May 20 '12 at 16:45
Use $\epsilon$-type definition for limit. – GEdgar May 20 '12 at 16:49
The most pedestrian way to show this is to cut the sum in two parts and show that the first part is small because the $t$-factor is small and the second part is small because the $q$-factor is small. – Phira May 20 '12 at 16:50
@Phira: please illustrate - I have tried that approach to no avail. – akkkk May 20 '12 at 16:51
up vote 1 down vote accepted

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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Okay, so what is a suitable large $n$? – akkkk May 20 '12 at 17:00
We need that $(t_i)$ is small when $i>n/2$, this gives one condition on $n$. Also, $\sum_{i=n/2+1}^\infty q^i$ should be small, which gives another condition. – Eepzy May 20 '12 at 17:13
I'd like a more formal proof - how do I pick $N$ such that $|\sum_{i=0}^n q^i t_{n-i}|<\epsilon$ for all $n\geq N$? – akkkk May 20 '12 at 17:42
Pick $N$ such that $|t_i|<\epsilon$ and $\left|\sum_{j=i}^\infty q_j\right|<\epsilon$ for $i>N/2$. Then use the splitting argument discussed above to obtain $\left|\sum_{i=0}^n q^i t_{n-i}\right|\leq C_1 \epsilon + C_2 \epsilon$, where $C_1$ is a bound on $\left|\sum_{i=0}^K q_i\right|$ and $C_2$ on $|t_i|$. You can get rid of the constants by replacing $\epsilon$ by $\epsilon/(2C_i)$ in the first sentence. – Eepzy May 20 '12 at 17:57
$C_1=\sup_K \left| \sum_{i=0}^K q_i\right|$, $C_2=\sup_i |t_i|$. These are finite since $(t_i)$ and $\sum_{i=0}&\infty q_i$ are convergent. They are fixed numbers and do not depend on $n$. – Eepzy May 20 '12 at 18:09

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