Show that $\sum_{n=0}^\infty |a_n|<\infty.$ 
Let $\{a_n\}$ be a real sequence. Suppose for every convergent real sequence $\{s_n\}$, the series $\sum_{n=0}^\infty a_n s_n$ converges. Show that $\sum_{n=0}^\infty |a_n|<\infty.$

Mt attempt:
Consider the sequence $\{s_n\}$ where $s_n=1$ for all $n$. Then we have that $\sum_{n=0}^\infty a_n$ converges. But how do I show the absolute convergence? Do I have to do something using Abel summation here? Because since $\sum_{n=0}^\infty a_n$ converges, let $\sum_{n=0}^\infty a_n=s$. Then the Abel sum $\lim_ {x\rightarrow 1} \sum_{n=0}^\infty a_nx^n=s$. I am stuck here. Can somebody please help me?
 A: If $\sum_{n=0}^{\infty} |a_n| = \infty,$ we can find $0=n_1 < n_2 < \cdots $ such that
$$\sum_{n=n_k}^{n_{k+1}-1}|a_n| > k.$$
For each $n\in \mathbb N,$ there is a unique $k = k(n)$ such that $n_k\le n < n_{k+1}.$ Define $s_n = \text { sgn }(a_n)/k(n)$ for each $n.$ Then $s_n \to 0,$ hence $\sum a_ns_n$ converges. But
$$\sum_{n=1}^{\infty}a_ns_n = \sum_{k=1}^{\infty}\sum_{n_k\le n < n_{k+1}}\frac{|a_n|}{k} > \sum_{k=1}^{\infty}1 = \infty.$$
That is a contradiction, giving the result.
A: Just elaborating Patrick's comment. Fix $N$. Let 
$$
s_n =\begin{cases}
|a_n|/a_n &\text{if } n\leq N \text{ and } a_n\neq 0 \\
1  &\text{if } n\leq N \text{ and } a_n=0 \\
1 &\text{if } n> N
\end{cases}
$$
It is clear that $s_n$ is a convergent sequence. So by the hypothesis, 
$$
\sum_{n=1}^{\infty} s_n a_n = \sum_{n=1}^{N} |a_n| + \sum_{n=N+1}^{\infty} a_n
$$
converges. Now let 
$$
t_N = \sum_{n=1}^{N} |a_n| + \sum_{n=N+1}^{\infty} a_n
= \sum_{n=1}^{\infty} |a_n| - \sum_{n=N+1}^{\infty} |a_n| + \sum_{n=N+1}^{\infty} a_n
$$
We have shown that $t_N$ is well-defined for each $N$. Now note that
$$
\lim_{N\to\infty} \sum_{n=N+1}^{\infty} \left(a_n-|a_n|\right) \leq \lim_{N\to\infty}  \sum_{n=N+1}^{\infty} \left(2 a_n \right) = 0
$$
because $\sum_{n=1}^{\infty} a_n$ converges, as you pointed out in the post. We have
$$
t_N = \sum_{n=1}^{\infty} |a_n|  + p_N
$$
where $p_{N} = \sum_{n=N+1}^{\infty} (a_n-|a_n|)$. Since $p_{N}\to 0$, we can find a sufficiently large $N$ such that $|p_{N}|<1$. But then
$$
\sum_{n=1}^{\infty} |a_n| = t_{N} - p_{N} < t_{N} + 1
$$ 
so in particular $\sum_{n=1}^{\infty} |a_n|$ converges because $t_{N}+1$ is a finite number. 
