Let $(a_n)_{n \in\ \mathbb{N^*}}(b_n)_{n \in\ \mathbb{N^*}}$ sequences of real numbers. Show that:

If $\sum\limits_{n=1}^{\infty}a_n$ is bounded and if $(b_n)_{n \in\ \mathbb{N^*}}$ converges monotone to $0$, then $\sum\limits_{n=1}^{\infty}a_nb_n$ converges.

I know, i have to use the Leibniz's test and i have to use, that $\sum\limits_{k=1}^{n}a_kb_k = A_nb_{n+1}+\sum\limits_{k=1}^{n}A_k(b_k-b_{k+1})$ is for $A_k=\sum\limits_{j=1}^{k}a_j$ for $1 \leq k \leq n$

Futhermore, i've already proofed that, if $\sum\limits_{n=1}^{\infty}a_n$ converges absolutely and if $(b_n)_{n \in\ \mathbb{N^*}}$ is bounded, then $\sum\limits_{n=1}^{\infty}a_nb_n$ converges absolutely.


We have to prove that the sequence $(S_N=\sum_{n=1}^Na_nb_n)_{N\geqslant 1}$ is Cauchy. Using the partial summation, we get $$S_{M+N}-S_N=A_{M+N}b_{M+N+1}-A_Nb_{N+1}+\sum_{k=N+1}^{N+M}A_k(b_k-b_{k+1}).$$ The first two terms go to $0$ as $M,N\to \infty$ since $(A_N)_{N\geqslant 1}$ is bounded, and for the third, we use the bound $$\left|\sum_{k=N+1}^{N+M}A_k(b_k-b_{k+1})\right|\leqslant \sup_{j\in\mathbb N}|A_j|\cdot \sum_{k=N+1}^{M+N}(b_k-b_{k+1})=\sup_{j\in\mathbb N}|A_j|\cdot b_{N+1}.$$


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