Let $\{a_k\}_{k\leq 1}$ be a sequence of real numbers such that $$\sum_{k=1}^{\lfloor X \rfloor} a_k = O(X^a),$$ for some $a>0$, and real values $X>0$. Now let $\{b_k\}_{k\geq 1}$ be a sequence of positive numbers (the modifiers) such that $b_k = f(k)>0$ where $f(X)=O(X^{-b})$ for some $b>0$, and if necessary, we can assume that $f$ is decreasing as $X$ increases.
My questions is what can one say about the growth of $$\sum_{k=1}^{\lfloor X \rfloor} a_k b_k.$$ In particular, is it possible to show that $\sum_{k=1}^{\lfloor X \rfloor} a_k b_k = O(X^{a-b})$?
If you do not assume $f$ to be decreasing, it is hopeless. Give sequences like $1,-1,1/2^a,-1/2^a,1/3^a,-1/3^a...$, $1,0,1/2^b,0,1/3^b,0...$, resp, then you see the sum is same in scale with $\sum_{k=1}^{X} k^{a-b}$ then all we can say is $O(X^{a-b+1})$ if $a\neq b$, and $O(X\log X)$ if $a=b$.
You can easily show this using $a_n=O(n^a)$.
Assume $f$ is decreasing.
We can rewrite the sum as $\sum_{k=1}^n A_k(b_k-b_{k+1})+A_nb_{n+1}$, where $A_n=\sum_{k=1}^n a_k$.
Then, since $b_k-b_{k+1}>0$, it suffices to consider only when $\forall k,A_k=k^a$.
Now, the sum is bounded above by $\sum k^{a-b-1}$ and that is $O(X^{a-b})$ if $a\neq b$, and $O(X^{a-b}\log X)$ if $a=b$.
That these bounds are optimal can be checked via putting $\forall k,a_k=k^a-(k-1)^a,b_k=k^{-b}$.