Let $\left\{a\_k\right\}\_{0\leqslant k\leqslant n}$ and $\left\{b\_k\right\}\_{0\leqslant k\leqslant n}$ be two positive decreasing sequences such that $c\_1a\_k\leqslant b\_k\leqslant c\_2 a\_k$ for all $k$ for some positive constants $c\_1,c\_2$. Let $\epsilon>0$ and $f>0$ a function such that $$\sum\_{k=0}^na\_k\sin(k\epsilon)\geqslant f(n).$$



> 
> **Question.** Is it true that $$\sum\_{k=0}^nb\_k\sin(k\epsilon)\geqslant cf(n)$$ for some constant
> $c$ (independent of $n$, $\epsilon$)?
> 
> 
> 




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Without the monotonicity condition on the $a\_k$'s and $b\_k$'s, the answer is no, as shown in [this answer](https://mathoverflow.net/a/390785/36721).



Oscillating sums II
$\newcommand\ep\varepsilon$The answer is still no (assuming you wanted $c>0$). Indeed, let
$$s\_n:=\sum\_{k=0}^n a\_k\sin(k\ep),\quad t\_n:=\sum\_{k=0}^n b\_k\sin(k\ep),$$
where $\ep=3\pi/4$, $a\_0=b\_0=10$, $a\_1=1$, $b\_1=9$, and $a\_k=b\_k=2^{-k}$ for $k\ge2$.


Then $c\_1a\_k\le b\_k\le c\_2a\_k$ for $c\_1=1$, $c\_2=9$, and all $k\ge0$. Moreover, $s\_0>0$ and $s\_1=-1 + 5\sqrt2>6>0$, whence $s\_n>6-\sum\_{k=2}^\infty 2^{-k}>0$ for $n\ge2$. So, $s\_n>0$ for all $n\ge0$. So, letting
$f(n):=s\_n$ for all $n\ge0$, we get a positive function $f$ such that $s\_n\le f(n)$ for all $n\ge0$. Thus, all your conditions on the $a\_k$'s and $b\_k$'s are satisfied.


However, $t\_1=-9 + 5\sqrt2<-19/20<0$ and, moreover, $t\_n<-19/20+\sum\_{k=2}^\infty 2^{-k}<0$ for all $n\ge2$. Thus, the condition $t\_n\le cf(n)$ fails to hold for any $c>0$ and any $n\ge1$.


