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Prove that if $\lim\limits_{n\to\infty}a_n=g$ and $\lim\limits_{n\to\infty}(b_1+b_2+\cdots+b_n)=\infty$ wherein constantly $b_n\geq0$, then $$\lim\limits_{n\to\infty}\dfrac{a_1b_1+a_2b_2+\cdots+a_nb_n}{b_1+b_2+\cdots+b_n}=g$$ I'd like to see a proof to this exercise because in my textbook there are no answers and I find it hard

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Denote the numerator as $A_{n}$ and denominator as $B_{n}$.

Since,

$$\displaystyle \lim_{n \to \infty} \left(\dfrac{A_{n+1} - A_{n}}{B_{n+1} - B_{n}}\right) = \lim_{n \to \infty} \dfrac{a_{n+1}b_{n+1}}{b_{n+1}} = g$$

we have, by the Stolz–Cesàro Theorem,

$$ \lim_{n \to \infty} \dfrac{A_{n}}{B_{n}} = g $$

as desired.

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