# Norm of linear functional in $l_1$

I'd like some help to solve the following question:

Let $(a_i)$ be a sequence in $\mathbb{R}$ such that $\sum_{i=1}^\infty a_ix_i< \infty$ for all $(x_i)\in l_1(\mathbb{R})$. Show that $(a_i)\in l_\infty$.

My attempt. For every $n\in \mathbb{N}$, set $f_n:l_1\rightarrow \mathbb{R}$ as $f_n(x)=\sum_{i=1}^na_ix_i$, where $x=(x_i)$. Clearly, $f_n$ is a linear functional in $l_1$. Moreover, since $$|f_n(x)|\leq \sum_{i=1}^n|a_ix_i|\leq \max_{1\leq i\leq n}|a_i| \cdot \|x\|_1,$$ $f_n$ is bounded. I know if I get $\|f_n\|= \max_{1\leq i\leq n}|a_i|$, I can use the uniform boundedness principle to obtain the result. However, I couldn't find an element $x\in l_1$ such that $$\frac{|f_n(x)|}{\|x\|_1}= \max_{1\leq i\leq n}|a_i|.$$ I'll be grateful for any hint. Thanks in advance!

Pick $k \leqslant n$ such that $\lvert a_k\rvert = \max_{1 \leqslant i \leqslant n} \lvert a_i\rvert$. Then
$$\lvert f_n(e_k)\rvert = \lvert a_k\rvert = \max_{1 \leqslant i \leqslant n} \lvert a_i\rvert \cdot \lVert e_k\rVert_1$$
for $e_k$ the $k^{\text{th}}$ standard unit vector, i.e.
$$(e_k)_i = \begin{cases} 1 &\text{if } i = k \\ 0 &\text{if } i \neq k \end{cases}\,.$$