Is $\sum_{i = 1}^\infty a_i \le \sum_{i = 1}^\infty b_i$ if $a_i \le b_i$? my question is: Is $\sum_{n = 1}^\infty a_n \le \sum_{n = 1}^\infty b_n$ if $a_n \le b_n$?
In my case I have to prove 
$$\sum_{n = 1}^\infty a_n \overline{b_n}~~~\text{converges absolutely} \Leftrightarrow \sum_{n = 1}^\infty |a_n \overline{b_n}| < \infty $$
$\forall a_n,b_n\in l^2(\mathbb{C})= \left\{(a_n)_{n\in\mathbb{N}} ~~\big| ~~ a_n \in \mathbb{C} ~~ \land  ~~ \sum_{n = 0}^\infty |a_n|^2 < \infty \right\}$
I wanted to use the relation $|ab| \le \frac{1}{2}(|a|^2+|b|^2)$ with $a,b, \in \mathbb{C}$. And a friend of mine suggested doing that the following way:
$$\sum_{n = 1}^\infty |a_n \overline{b_n}| = \sum_{n = 1}^\infty |a_n b_n| \le \sum_{n = 1}^\infty \frac{1}{2}(|a_n|^2+|b_n|^2)$$  
Is that possible?
I think not - because we are working with sums there and not with just the sequence.
Thank you very much for your help.
FunkyPeanut
 A: Suppose that $a_n\leq b_n$ for any $n\in\mathbb N$. Assume that both $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ exist (otherwise, the question simply doesn't make sense). It is not difficult to show that in this case, the sum $\sum_{n=1}^{\infty} (a_n-b_n)$ exists, too, and is equal to $\sum_{n=1}^{\infty} a_n-\sum_{n=1}^{\infty} b_n$. Now, since $a_n\leq b_n$ for all $n\in\mathbb N$, it follows that for any $N\in\mathbb N$, $\sum_{n=1}^N(a_n-b_n)\leq 0$. Taking the limit,
$$\sum_{n=1}^{\infty} a_n-\sum_{n=1}^{\infty} b_n=\sum_{n=1}^{\infty}(a_n-b_n)\underset{\text{def}}{=}\lim_{N\to\infty}\sum_{n=1}^N(a_n-b_n)\leq0,$$
which is what we needed to show.
Added: If $b_n\geq a_n\geq 0$ for all $n\in\mathbb N$, then that $\sum_{n=1}^{\infty} b_n$ exists implies also that $\sum_{n=1}^{\infty} a_n$ exists, too. This is because
$$0\leq\sum_{n=1}^{N}a_n\leq\sum_{n=1}^{N}b_n\leq\lim_{N\to\infty}\sum_{n=1}^{N}b_n=\sum_{n=1}^{\infty}b_n<\infty\quad\forall N\in\mathbb N.$$
Therefore the sequence $(\sum_{n=1}^{N}a_n)_{N\in\mathbb N}$ is monotonically increasing and bounded above, so its limit $\sum_{n=1}^{\infty}a_n$ exists. In addition, since $\sum_{n=1}^{N}a_n\leq\sum_{n=1}^{\infty}b_n$ for all $N\in\mathbb N$, this inequality is preserved in the limit: $\sum_{n=1}^{\infty}a_n\leq\sum_{n=1}^{\infty}b_n.$
If you combine these results, you have everything you need to prove your main claim.
