This proposition was told to me by my senior, and it's been bothering me for a long time. Now, I can only prove that this proposition is true when $\left \{ a_n \right \}$ is a non-negative sequence. The proposition is as follows:
Let $\left \{ b_n \right \}$ be a monoton-decreasing sequence of positive number and $\left \{ a_n \right \} \subseteq \mathbb{R}$. If $\sum\limits_{n=1}^{\infty} a_n b_n $ converges and $\lim\limits_{n \to \infty} b_n=0$, then we have
$$
\lim_{n \to \infty} \left ( a_1+a_2+…+a_n \right ) b_n=0
$$
Here is my some reserch of this problem. If $a_n\ge 0$, I can use this method to prove.
First of all, we note that
$$
b_1=b_1-\lim_{n \to \infty} b_n=b_1-\left ( \sum_{n=1}^{\infty}b_{n+1}-b_{n}+b_1 \right ) =\sum_{n=1}^{\infty}b_{n}-b_{n+1}
$$
Hence we can see $b_1$ as a positive series, and we can consider $b_n$ as the remainder of the series $\sum\limits _{n=1}^{\infty} b_n-b_{n+1}$, which is
$$
b_n=\sum_{k\ge n}^{\infty} b_k-b_{k+1}
$$
Then we have
$$
\sum\limits_{n=1}^{\infty} a_nb_n =\sum\limits_{n=1}^{\infty}a_n\sum\limits_{k\ge n}^{\infty}(b_{k}-b_{k+1})=\sum\limits_{n=1}^{\infty}\sum\limits_{k\ge n}^{\infty}a_n(b_{n}-b_{n+1})
$$
Since $\sum\limits_{n=1}^{\infty}\sum\limits_{k\ge n}^{\infty}a_n(b_{k}-b_{k+1})$ is positive series, then we can rearrange the order of the term of series, and we obtain
$$
\sum\limits_{n=1}^{\infty}\sum\limits_{k\ge n}^{\infty}a_n(b_{k}-b_{k+1})=\sum\limits_{n=1}^{\infty} A_n(b_{n}-b_{n+1})
$$
On the other hand, note that performing the Abel transformation on $\sum\limits_{p=1}^{n} a_pb_p$ yields
$$
\begin{aligned}
b_nA_n=\sum_{p=1}^{n} a_pb_p+\sum_{p=1}^{n-1} (b_{p+1}-b_p)A_p\\
=\sum_{p=1}^{n} a_pb_p-\sum_{p=1}^{n-1} (b_p-b_{p+1})A_p
\end{aligned}
$$
where $A_n=\sum\limits_{p=1}^{n}a_p$. When $n\to \infty$, we have
$$
\lim_{n \to \infty} b_nA_n=\sum_{p=1}^{\infty } a_pb_p-\sum_{p=1}^{\infty} (b_p-b_{p+1})A_p=0
$$
Now, we have proven the case of $a_n\ge 0$. In addition, during the exploration of this question, I also discovered an identities such as
$$
\sum\limits_{j=1}^{n} A_j(b_{j+1}-b_{j})=\sum\limits_{j=1}^{n} a_j(b_{j}-b_{n})
$$
But when it comes to proving the general situation, I seem to have fallen into a very strange cycle, and I can't escape the cycle. I don't know how to combine the two conditions of sequence monotonicity and series convergence.
Best Answer
This is Kronecker's lemma in disguise.
Let $S_n = \sum_{k=1}^{n} a_k b_k$ with the understanding that $S_0 = 0$ and $S_n \to S_{\infty} \in \mathbb{R}$. Then by summation by parts,
\begin{align*} \sum_{k=1}^{n} a_k &= \sum_{k=1}^{n} \frac{a_k b_k}{b_k} = \frac{S_n}{b_n} - \sum_{k=1}^{n-1} \left( b_{k+1}^{-1} - b_k^{-1} \right) S_k. \end{align*}
Multiplying $b_n$ to both sides,
\begin{align*} \left( \sum_{k=1}^{n} a_k \right) b_n &= S_n - \frac{\sum_{k=1}^{n-1} \left( b_{k+1}^{-1} - b_k^{-1} \right) S_k}{b_n^{-1}}. \end{align*}
Now by the Stolz–Cesàro theorem, we have
$$ \lim_{n\to\infty} \frac{\sum_{k=1}^{n-1} \left( b_{k+1}^{-1} - b_k^{-1} \right) S_k}{b_n^{-1}} = \lim_{n\to\infty} \frac{\left( b_{n}^{-1} - b_{n-1}^{-1} \right) S_{n-1}}{b_n^{-1} - b_{n-1}^{-1}} = S_{\infty}. $$
Hence it follows that
\begin{align*} \lim_{n\to\infty} \left( \sum_{k=1}^{n} a_k \right) b_n = S_{\infty} - S_{\infty} = 0. \end{align*}