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Suppose $\{a_n\}$ and $\{b_n\}$ satisfy
(i):$\;a_1+a_2+\cdots+a_n$ is bounded for all $n$.
(ii):$\;\lim b_n=0$.
(iii):$\;|b_1-b_2|+|b_2-b_3|+\cdots+|b_n-b_{n+1}|+\cdots<∞.$
Prove that $Σa_nb_n$ is convergent.

I find it's like Dirichlet discriminant method,but I can't prove it.

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Denote $S_n=\sum_{k=1}^n a_k$ with $S_0:=0$, then $|S_n|\leqslant M <\infty$ for all $n$ by assumption. We have for $m<n$, \begin{align*} \Big|\sum_{k=m}^n a_kb_k\Big|&=\Big|\sum_{k=m}^n (S_k-S_{k-1})b_k\Big|=\Big|\sum_{k=m}^n S_k b_k -\sum_{k=m}^n S_{k-1}b_k\Big|\\ &=\Big|S_{m-1}b_m+\sum_{k=m}^{n-1} S_k(b_k-b_{k+1})+S_n b_n\Big|\leqslant |S_{m-1}b_m|+\sum_{k=1}^{n-1} |S_k|\cdot|b_k-b_{k+1}|+|S_n b_n|\\ &\leqslant M|b_m|+M \sum_{k=m}^{n-1}\cdot|b_k-b_{k+1}| + M|b_n|. \end{align*} Taking $m, n\to \infty$ both sides, we get $$\Big|\sum_{k=m}^n a_kb_k\Big| \to 0,$$ hence the we get the convergence of series.

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