I'm trying to understand a proof for the following problem:

If $\sum_{n=0}^{\infty} a_n$ is absolutely convergent and $\lim b_n=0$, take $c_n=a_0 b_n + a_1 b_{n-1} + \cdots + a_n b_0$ and prove that $\lim c_n=0$.

The proof I'm trying to understand follows:

Take $|b_n| \leq B$ for all $n \geq 0$ and $\sum |a_n|=A$. Given an $\varepsilon>0$, there is a $n_0 \in \mathbb{N}$ such that $n \geq n_0 \Rightarrow |b_n|< \varepsilon/2A$ and $|a_n| + |a_{n+1}|+ \cdots < \varepsilon/2B$. So $n>2n_0 \Rightarrow$

$$|c_n|=|a_0 b_n + \cdots + a_{n_0} b_{n-n_0}+a_{n_0+1}b_{n-n_0-1}+\cdots+a_n b_0|$$

$$\leq (|a_0|+\cdots+|a_{n_0}|)\frac{\varepsilon}{2A}+(|a_{n_0+1}|+\cdots+|a_n|)\cdot B$$


Therefore, $\lim c_n = 0$.

The part that I don't understand is where it states that $|a_n| + |a_{n+1}|+ \cdots < \varepsilon/2B$. How can it be true if this series can converge to any positive value and I can pick smaller and smaller $\varepsilon$'s (and I can also pick bigger and bigger B's) which would make $|a_n| + |a_{n+1}|+ \cdots > \varepsilon/2B$. I know I must be getting something wrong out of this detail, so any help for me to understand it would be much appreciated.

  • $\begingroup$ hm, at the end all must be lower than $\epsilon>0$ $\endgroup$ – Dr. Sonnhard Graubner Jun 29 '17 at 16:57

If you make $\epsilon$ smaller or $B$ bigger, the $n_0$ will be bigger as well. For each value of $\frac{\epsilon}{2B}$ there exists a $n_0$, this is not necessarily the same for each $\frac{\epsilon}{2B}$. So the important part is that the subscripts in $|a_n| + |a_{n + 1}| + ...$ are not fixed.


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