This question was posted/originated after a failure of a more generic attempt here:
Let $\alpha_n$ be a sequence of positive Reals. It is known that $$\alpha_n \sim \log(n)$$
Let $\beta_n$ be another sequence of positive Reals such that, $$\sum_{k = 1}^n\beta_k \sim \log(n)$$ Can we say/prove that $$\frac{\sum_\limits{k = 1}^n\alpha_k\beta_k }{\sum_\limits{k = 1}^n\beta_k} \sim \frac{1}{2}\log(n)$$
PS : $\alpha_n \sim \log(n)$ is equivalent to saying $\lim_\limits{n\to\infty}\frac{\alpha_n}{\log(n)} = 1$
Let $B(x)=\sum_{n\leq x} \beta_n$. Then $B(x)\sim \log x$.
For any $\epsilon>0$, there exists $m>0$ such that $$ n\geq m \Longrightarrow (1-\epsilon)\log n < \alpha_n < (1+\epsilon)\log n, \ \ \textrm{and} $$ $$ t\geq m \Longrightarrow (1-\epsilon)\log t<B(t)<(1+\epsilon) \log t. $$
Fon any $N>m$, $$ (1-\epsilon) \sum_{n=m+1}^N \beta_n\log n < \sum_{n=m+1}^N \alpha_n\beta_n < (1+\epsilon)\sum_{n=m+1}^N \beta_n\log n $$
By Abel's summation formula, we have $$ \sum_{m<n\leq N} \beta_n\log n = B(N)\log N-B(m)\log m - \int_m^N \frac{B(t)}t dt $$
We have $B(N)\log N \sim (\log N)^2$. For the integral, we have $$ (1-\epsilon) \int_m^N \frac{\log t}t dt \leq \int_m^N \frac{B(t)}tdt \leq (1+\epsilon)\int_m^N \frac{\log t}t dt. $$ Therefore, taking $N\rightarrow\infty$, we obtain that $$ \limsup_{N\rightarrow\infty}\frac1{(\log N)^2} \sum_{m<n\leq N} \beta_n\log n \leq 1- \frac{1-\epsilon}2 =\frac{1+\epsilon}2, \ \ \textrm{and} $$ $$ \liminf_{N\rightarrow\infty}\frac1{(\log N)^2} \sum_{m<n\leq N} \beta_n\log n \geq 1- \frac{1+\epsilon}2 =\frac{1-\epsilon}2. $$ Letting $\epsilon\rightarrow 0$, we have the result: $$ \lim_{N\rightarrow\infty}\frac1{(\log N)^2} \sum_{m<n\leq N}\beta_n\log n = \frac12. $$ Now, we can go back to $\sum \alpha_n\beta_n$: $$ (1-\epsilon)\frac1{(\log N)^2} \sum_{m<n\leq N} \beta_n \log n < \frac1{(\log N)^2} \sum_{m<n\leq N}\alpha_n\beta_n <(1+\epsilon) \frac1{(\log N)^2}\sum_{m<n\leq N}\beta_n\log n $$ Similarly taking $\limsup_{N\rightarrow\infty}$, $\liminf_{N\rightarrow\infty}$, and $\epsilon\rightarrow 0$, we obtain the result: $$ \lim_{N\rightarrow\infty}\frac1{(\log N)^2} \sum_{m<n\leq N} \alpha_n\beta_n = \frac12. $$ This will be also true if we replace $m$ by $0$.