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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?

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    $\begingroup$ Let $s_n$ be the sign of $a_n$ for the first $N$ terms and $1$ after that. Now let $N$ go to infinity. $\endgroup$ – user4571 Jan 26 '16 at 3:30
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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.

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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.

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  • $\begingroup$ But, where did you prove that $\lim t_N$ is finite? $\endgroup$ – Extremal Jan 26 '16 at 3:54
  • $\begingroup$ @EpsilonDelta I have now edited my answer. Thanks for the heads-up, and I hope the argument is simpler now. $\endgroup$ – Prism Jan 26 '16 at 4:09
  • $\begingroup$ @Prism: Why is $\sum (|a_n|-a_n )< \sum 2 a_n$? Also you say $t_N$ is well-defined after you relate it to $\sum_{n=1}^{\infty} |a_n|$. So you are using what you are trying to prove, i.e., $\sum_{n=1}^{\infty} |a_n| < \infty$ $\endgroup$ – RRL Jan 26 '16 at 4:33
  • $\begingroup$ @RRL: I say $t_N$ is well-defined because we have proved it to be! I don't think there is a circular logic here, but perhaps my presentation is poor. Also, thanks for the catch. It should be $\sum (a_n - |a_n|) \leq \sum 2a_n$ because if $a_n$ is positive, then $a_n-|a_n|=0$, and if $a_n$ is negative, then $a_n-|a_n|=2a_n$. Thus, $\sum (a_n-|a_n|)\leq \sum 2a_n$. I will edit the answer accordingly. $\endgroup$ – Prism Jan 26 '16 at 15:02
  • $\begingroup$ @RRL: I know $t_N$ is well-defined because the series it represents converges. Writing $\sum_{n=1}^{\infty} |a_n|$ inside $t_{N}$ is just algebraic manipulation. I am not assuming that $\sum_{n=1}^{\infty} |a_n|$ is finite there. I think there ought to be a better write-up of this approach that doesn't make it seem circular. $\endgroup$ – Prism Jan 26 '16 at 15:07

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