Proving limit of absolute convergent sequence is larger than limit of convergent sequence

Question

I have a bounded sequence $A_{n}=\sum\limits_{k=1}^{n}|x_{k}|$ and a bounded sequence $B_{n}=\sum\limits_{k=1}^{n}x_{k}$.

Now I've already proved both sequence converge but now I'm left to prove that the following holds for the limit $B$ of $B_{n}$ and $A$ of $A_{n}$: $|B| \leq A$.

I'm trying to use the following proposition: assume that $\lim_{n\to\infty} x_{n}=L$. If $x_{n} <a$ for some $a \in \mathbb{R}$ and all $n \in \mathbb{N}$ then also $L \leq a$.

Now I don't know how to proceed. Clearly $B_{n} \leq A_{n}$ for all $n \in \mathbb{N}$, but this doesn't give us a clear $a$ from the proposition.

Any suggestions?

Answer 1

Since $(\forall n\in\mathbb N):\lvert B_n\rvert\leqslant A_n$ and since the absolute value function is continuous,\begin{align}\lvert B\rvert&=\left\lvert\lim_{n\to\infty}B_n\right\rvert\\&=\lim_{n\to\infty}\lvert B_n\rvert\\&\leqslant\lim_{n\to\infty}A_n\\&=A.\end{align}