Let $\alpha_n$ be a sequence of positive Reals. It is known that $$\alpha_n \sim g(n)$$ where $g(n)$ is a strictly increasing function.
Let $\beta_n$ be another sequence of positive Reals such that, $$\sum_{k = 1}^n\beta_k \sim g(n)$$ Can we say/prove that $$\frac{\sum_\limits{k = 1}^n\alpha_k\beta_k }{\sum_\limits{k = 1}^n\beta_k} \sim \frac{1}{2}g(n)$$
PS : $\alpha_n \sim g(n)$ is equivalent to saying $\lim_\limits{n\to\infty}\frac{\alpha_n}{g(n)} = 1$
I do not believe this claim is true, at least without additional conditions on $g(n)$.
Define $\alpha_n:=\sum_{k=1}^n \beta_k$, where $\beta_k$ is an arbitrary sequence of positive real numbers. (These sequences now clearly satisfy the hypotheses of the claim.)
Then \begin{eqnarray} & &\frac{\sum_{k=1}^n \alpha_k\beta_k}{\sum_{k=1}^n \beta_k}\\ &=&\frac{\sum_{k=1}^n \beta_k\sum_{j=1}^k\beta_j}{\sum_{k=1}^n \beta_k}\\ &=& \frac{\left(\sum_{k=1}^n \beta_k\right)^2+\sum_{k=1}^n \beta_k^2}{2\sum_{k=1}^n \beta_k}\\ &=& \frac{1}{2}\sum_{k=1}^n \beta_k+\frac{\sum_{k=1}^n\beta_k^2}{2\sum_{k=1}^n \beta_k}. \end{eqnarray} So in this particular case, the claim is equivalent to the assertion that
\begin{equation} \frac{\sum_{k=1}^n \beta_k^2}{\left(\sum_{k=1}^n \beta_k\right)^2}\rightarrow 0. \end{equation}
This need not be true if $\beta_k$ grows rapidly, as can be explicitly seen by evaluating the limit for $\beta_k:=\exp(k)$.