Question on the inequality of sequences Given two sequence $(a_n)_{n \geq 0}$ , $(b_n)_{n \geq 0}$ satisfing $a_n,b_n >0$ for all $n$ and $\sum_{n}a_n \gtrsim \sum_{n}b_n$. My question is that: For a sequence $(c_n)_{n \geq 0}$ be positive, we have the following inequality ?
$\sum_{n}a_nc_n \gtrsim \sum_{n}b_nc_n$
 A: Let 
$$
a_n = 
\begin{cases}
n &\ \text{if}\ n\ \text{is odd}\\
n^{-3} &\ \text{if}\ n\ \text{is even}\\
\end{cases}
$$
$$
b_n = 
\begin{cases}
n^{-2} &\ \text{if}\ n\ \text{is odd}\\
1 &\ \text{if}\ n\ \text{is even}\\
\end{cases}
$$
$$
c_n = 
\begin{cases}
n^{-4} &\ \text{if}\ n\ \text{is odd}\\
1 &\ \text{if}\ n\ \text{is even}\\
\end{cases}
$$
Then $\sum_{n = 1}^Na_n \sim N^2 \ge N \sim\sum_{n = 1}^Nb_n$, but $\sum_{n = 1}^Nb_nc_n \sim N \gtrsim \sum_{n = 1}^Nn^{-3} = \sum_{n = 1}^Na_nc_n$.
A: The answer is negative. Consider the three sequences: 
$$
(a_n)= (1,0,0\ldots) \quad (b_n)=\left( 0, \frac12, 0\ldots \right),\quad (c_n)=\left(\frac{1}{4}, 1, \ast, \ast \ldots\right).$$
(Here $\ast$ means any positive number). You have that 
$$
\sum_n a_n=1>\frac12=\sum_n b_n, $$
but
$$
\sum_n c_na_n=\frac14<\frac12=\sum_nc_nb_n.$$
P.S.: I noticed that you require $a_, b_n >0$, so strictly speaking the present example is ruled out. But it can be fixed easily, replacing the zeroes with any converging series. 
