Showing $(a_k)_{k=1}^{\infty}$ is in $l^1$ My lecturer claimed this following without a proof. If anybody could give me some idea why this is true, it would be great.
Let $a_k\in\mathbb{R}$, be a real sequence, such that the series $\sum_{k=1}^{\infty}a_kx_k$ converges for all sequences $(x_k)_{k=1}^{\infty}$ with $\lim_{k\rightarrow\infty}x_k=0.$ Then $(a_k)_{k=1}^{\infty}\in l^1$?
 A: It suffices to show that $(a_n)_{n=1}^\infty\in (c_0)^*$. To show this, observe that the operators $x\mapsto \sum_{n=1}^N a(n)x(n)$ are continuous on $c_0$. And since for each fixed $x\in c_0$, the quantity $$\sup_N |\sum_{n=1}^N a(n)x(n)|$$ is bounded. So by the Uniform Boundedness Theorem, $a$ is continuous on $c_0$.
A: We prove the opposite statement. Suppose $\displaystyle\sum\limits_{k=1}^\infty|a_k|=+\infty$. Consider the functionals $f_n(x)=\displaystyle\sum\limits_{k=1}^na_kx_k$, $x\in c_0$. It's easy to see that $f_n$ -- it's bounded linear functionals and $\|f_n\|=\displaystyle\sum\limits_{k=1}^n|a_k|$. By assuming $\sup\limits_n\|f_n\|=\displaystyle\sum\limits_{k=1}^\infty|a_k|=+\infty$. From  uniform boundedness principle we conclude that $\exists x_0\in c_0$ (i.e. $x_k^0\to0$) such that $\sup\limits_n|f_n(x_0)|=+\infty$, i.e. sequence $\displaystyle\sum\limits_{k=1}^na_kx_k^0$ is not bounded, what means it's not converge, thus the series $\displaystyle\sum\limits_{k=1}^\infty a_kx_k^0$ is not converge.
