Question regarding convergence to zero Let $0 \leq \epsilon < 1$ and $\{a_k\}$ be a sequence of positive numbers converging to $0$. What can we say about the sequence
$$b_n = \sum_{k=0}^n \epsilon^{n-k} a_k$$
It seems that this should converge to $0$.
 A: Your guess is correct. 
First of all, $b_n >0$ obviously. As $a_n$ is convergent, there is $M$ so that $a_n \le M$ for all $n$. On the other hand, let $\delta >0$. Then there is $N \in \mathbb N$ so that $a_n \le \delta$ for all $n \ge N$. Then for $m \ge N$, we have 
$$\begin{split}
b_m &= \sum_{k=0}^m \epsilon ^{m-k} a_k \\
&= \sum_{k=0}^{N-1} \epsilon ^{m-k} a_k+ \sum_{k = N}^m \epsilon ^{m-k} a_k\\
&\le M \sum_{k=0}^{N-1} \epsilon ^{m-k} + \delta \sum_{k = N}^m \epsilon ^{m-k} \\
&\le MN \epsilon^{m-N+1} + \delta \sum_{k=0}^m \epsilon^{m-k}\\
&\le MN\epsilon^{m-N} + \delta \frac{1-\epsilon^{m+1}}{1-\epsilon} \\
&\le  MN\epsilon^{m-N} + \delta \frac{1}{1-\epsilon}
\end{split}$$
Now for all $\gamma >0$, first we choose $\delta >0$ so that 
$$ \delta \frac{1}{1-\epsilon} <\frac{\gamma}{2},$$
then (fixing this $\delta$) we choose $N_\gamma > N$ so that 
$$ MN\epsilon^{m-N} < \frac{\gamma}{2}$$
whenever $m\ge N_\gamma$. Thus we have $b_m <\gamma $ whenever $m\ge N_\gamma$. Thus $\lim b_n =0$. 
