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?

  • 2
    $\begingroup$ Let $s_n$ be the sign of $a_n$ for the first $N$ terms and $1$ after that. Now let $N$ go to infinity. $\endgroup$
    – Potato
    Jan 26 '16 at 3:30

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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.