Question regarding convergence to zero

Question

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 $0$.

Answer 1

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 $\lim b_n =0$.