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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 $\lim_\limits{n\to\infty}\frac{\alpha_n}{g(n)} = 1$

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    where do you get the $1/2$ from? – 2017-02-01
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    have you tried to rewrite $2\alpha\beta=(\alpha+\beta)^2-\alpha^2-\beta^2$ – 2017-02-01
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    @tired No. Not that way. – 2017-02-01
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    I will say when its proved. – 2017-02-01

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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 $\beta_k:=\exp(k)$.

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    Thanks for the answer. I agree with you. In the problem I am dealing with, $$\sum_{k=1}^n\beta_k^2 \sim C$$, $C$ is a constant. Hope this helps to prove it. Atleast your example now is taken care of. Can I add this to the question with your permission. – 2017-02-01
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    Can you prove it for a specific $g(n)$? I am interested in the proof but not keen on keeping $g(n)$ generic. – 2017-02-01
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    If $\beta_k=2^{-k}$ for every $k$ then the numerator and denominator in the last expression in the A both have positive limits as $n\to \infty$. – 2017-02-01
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    @user254665 : Okay, Can you prove the result specifically for $g(n) = \log(n)$, Any ideas? – 2017-02-01
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    I'll think about it. I think you should pose it as a new Q, with a reference to this one. – 2017-02-01
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    @user254665 : posted : http://math.stackexchange.com/q/2124015/2987 – 2017-02-01