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Suppose $a_1,a_2,\ldots\in\mathbb{C}$ and $b_1,b_2,\ldots\in\mathbb{R}$.

Suppose also that $\sum a_n$ converges, that $b_n\leq b_{n+1}$ for all $n\geq 1$, and that $\lim_{n\rightarrow\infty}b_n=\infty$.

I want to show that

$$\lim_{N\rightarrow\infty}\dfrac{1}{b_N}\sum_{n=1}^Na_nb_n=0.$$

My intuition is that: As $N$ gets large, $a_n$ is very small (since its series is convergent), while $b_n$ for $n\leq N$ is bounded by $b_N$, which appears in the denominator, so $\dfrac{a_nb_n}{b_N}$ is small. How can we actually make the argument?

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Fix $\varepsilon\gt 0$: there is $M$ such that if $m\geqslant M$ and $n\geqslant 0$, then $$\left|\sum_{j=m+1}^{m+n}a_j\right|\lt\varepsilon.$$ Define for $k\geqslant M$: $s_k:=\sum_{j=M}^ka_j$. We have for $N\geqslant M$: $$\sum_{j=M+1}^Na_jb_j=\sum_{j=M+1}^N(s_j-s_{j-1})b_j=\sum_{j=M+1}^Ns_jb_j-\sum_{j=M}^{N-1}s_jb_{j+1}\\=s_Nb_N-s_Mb_{M+1}+\sum_{j=M+1}^{N-1}s_j(b_j-b_{j+1}),$$ hence $$\left|\frac 1{b_N}\sum_{j=1}^Na_jb_j\right|\leqslant \frac 1{b_N}\left|\sum_{j=1}^Ma_jb_j\right|+|s_N|+\frac{|s_M|\cdot |b_{M+1}|}{b_N}+\varepsilon\frac{b_N-b_{M+1}}{b_N}\\ \leqslant \frac 1{b_N}\left|\sum_{j=1}^Ma_jb_j\right|+2\varepsilon+\frac{|s_M|\cdot |b_{M+1}|}{b_N}.$$

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