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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.
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.
