So this is given: $\sum_{n=1}^{\infty}a_n*b_n $ converges for all sequences $(b_n)$, such that $\lim_{n \rightarrow \infty}b_n = 1$.

Somehow it should be showable that $(a_n)\,$converges absolutely, that is $\sum_{n=1}^{\infty}|a_n|$ converges.

I have been pondering with this for hours, and I appreciate any help and ideas.

  • 2
    $\begingroup$ Consider the sequence $a_n = \frac{(-1)^n}{n}$ then $\sum a_n b_n$ converges for all $b_n$ s.t. $\lim_{n\to \infty} b_n = 1$, but $\sum |a_n|$ does not converge. $\endgroup$ – Winther Nov 19 '14 at 1:28
  • $\begingroup$ We need some restriction on $(b_n)$, e.g. belonging to $\ell^p.$ $\endgroup$ – BigM Nov 19 '14 at 1:30
  • $\begingroup$ If you mean $\lim_{n\to \infty} |b_n| = 1$ then the proof is easy. Take $b_n = \text{sign}(a_n)$ then $|b_n|=1$, $a_nb_n = |a_n|$ and the result follows. $\endgroup$ – Winther Nov 19 '14 at 1:32
  • $\begingroup$ Winther: No it isn't the absolute value. Winther: How does that converge for all $b_n \rightarrow \infty$? $\endgroup$ – Mundo Nov 19 '14 at 1:37
  • 3
    $\begingroup$ your counter example is false. Just consider b_n=1+(-1)^n/log n. $\endgroup$ – Chen Jiang Nov 19 '14 at 2:09

Let me give a proof.

From the assumption, of course $\sum a_n$ converges. And also $\sum a_nb_n$ converges for all $\lim b_n=0$. Note here is $0$ since one can replace $b$ by $b-1$ to get convergence.

Now assume to the contrary that $\sum a_n$ does not converge absolutely.

Take $A_n=\max\{a_n,0\}$, the non-negative part and $B_n=-\min\{a_n,0\}$, the non-positive part. Then $\sum|a_n|=\sum A_n+\sum B_n=\infty$. By symmetry, we may assume that $\sum A_n$ is infinity (or just take the non-positive part instead). So we can find a increasing sequence of interger $n_k$ inductively such that $n_0=1$ and $$\sum_{n_k+1}^{n_{k+1}}A_n>k.$$ Now take $$b_n = \begin{cases} \frac{1}{k} & \text{ if } n_k<n\leq n_{k+1} \text{ and } A_n\neq 0;\\ 0 &\text{ if } A_n=0. \end{cases}$$ Then one can see that $\lim b_n=0$ but $$\sum a_nb_n=\sum A_nb_n=\sum_k\sum_{n_k+1}^{n_{k+1}}A_n\cdot\frac{1}{k}>\sum_k1=\infty.$$ A contradiction.

  • $\begingroup$ If $a_n$ does not converse absolutely, doesen't it follow that the negative part OR the non-negative part diverges. So $\sum A_n$ can in fact converge. And I cant get the last line, why $\sum a_nb_n=\sum A_nb_n$ ? $\endgroup$ – Mundo Nov 19 '14 at 14:54
  • 1
    $\begingroup$ edited. the sum are the same because $b_n=0$ if $a_n\neq A_n$ by the construction of $b_n$ $\endgroup$ – Chen Jiang Nov 19 '14 at 15:28

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