Define $S_N:=\sum_{k=1}^{N}a_k$, which is bounded. Let $(b_k)$ the decreasing sequence such that $\lim_{k\to \infty}b_k=0$. Show that $\sum_{k=0}^{\infty}a_kb_k$ converges.
I would like to have a feedback on my proof, please.
Proof.
Let $M>0$. As $\lim_{k\to \infty}b_k=0$, we have by definition that $\forall \epsilon>0 \ \exists n_0 \ \forall k\ge n_0$: $|b_k|<\epsilon/M$.
Then, as $S_N$ is bounded we have that $|S_N|\le M$ $\forall k$ such that $1\le k\le N$.
Suppose by absurd that $\sum_{k=0}^{\infty}a_kb_k$ diverges and so is not Cauchy:
$\exists \epsilon>0 \ \forall R \ \exists N \ge R$:
$|\sum_{k=R}^{N}a_kb_k|>\epsilon$.
But,
$|\sum_{k=R}^{N}a_kb_k|\le |\frac{\epsilon}{M}\sum_{k=R}^{N}a_k|\le \epsilon$. Contradiction. Therefore $\sum_{k=0}^{\infty}a_kb_k$ converges.
I doubt a little bit that it holds... It appears to me strange that i did't use the hypithesis that $(b_k)$ is decreasing. I think if we use this hypothesis, we obtain that: $0<b_{k+1}<b_k$ $\forall k\in \mathbf{N}$. But as i work with absolute values, i don't really understand where it could help me.
Moreover i think in my final inequality with the sum i should have decomposed it in the difference between two sums to be more rigorous... Thanks for help in advance!