Absolutely convergent series I am looking for an easy proof of the following theorem ( I know how to prove this by using Banach-Steinhaus), but I guess there must be something much easier:
If for every $(t_n)_n$ such that $t_n \rightarrow  0$ the series $\sum_n s_nt_n$ converges, then $\sum_n s_n$ converges absolutely.
Probably one needs to look at particular sequences $(t_n)$, but I do not know which one.
 A: Suppose that $\{s_n\}$ does not converge absolutely.  Begin with a sequence $r_n=\pm1$ such that $\sum_n r_ns_n=\infty$.  (ie $r_n=\frac{|s_n|}{s_n}$ if $s_n\neq0$, $r_n=1$ if $s_n=0$.)  We will now tamp the sequence $\{r_n\}$ down to a new sequence $\{t_n\}$ such that $t_n\to0$, but we still have $\sum _n t_n s_n$ diverges.
For each $N>0$, there is some $l_N>0$ such that $\displaystyle\sum_{n=0}^{l_n}r_ns_n>N^2$.  Choose $l_N$ greater if necessary so that $l_1<l_2<\cdots$.  Then for each $n$, we have $l_N\leq n<l_{N+1}$ for some $N$.  Define $t_n=\frac{r_n}{N}$.  It is clear now that $t_n\to0$.  Moreover, if you examine the partial sums $\displaystyle\sum_{n=1}^{l_N}t_ns_n>N$.  Thus the sum $\sum_nt_ns_n$ diverges.
A: I think an indirect proof may be the simplest. Suppose $\sum s_n$ does not converge absolutely. Let $n_0 = -1$, $M_0 = 0$, and for $k > 0$, let
$$n_k = \min \left\lbrace N : \sum_{n=1}^N \lvert s_n\rvert \geqslant M_{k-1}+1\right\rbrace;\quad M_k = \sum_{n=1}^{n_k} \lvert s_n\rvert.$$
For $n_{k-1} < n < n_k$, let $t_n = \frac1k\overline{\sigma(s_n)}$, where $\sigma(re^{i\varphi}) = e^{i\varphi}$ if $r > 0$, and $\sigma(0) = 1$.
Then $t_n \to 0$, and
$$\sum_{n = n_{k-1}+1}^{n_k} t_n s_n = \frac{1}{k} \sum_{n=n_{k-1}+1}^{n_k} \lvert s_n\rvert \geqslant \frac{1}{k},$$
so $\sum t_n s_n$ does not converge.
