If $a=\{a_n\}\in \ell^\infty(\mathbb{R})$ and $\langle a,x \rangle<\infty$ for all $x\in \ell^2(\mathbb{R})$, (where $\langle a, x\rangle=\displaystyle \sum_{k=1}^\infty a_kx_k$), then is $a\in \ell^2$?
So, $a$ defines a linear operator $T_a$ on $\ell^2$ by $T_a(x)=\langle a,x\rangle$. If the operator is bounded, then $a\in \ell^2$, since $\ell^2$ is a Hilbert space. But I don't see if $T_a$ should necessarily be bounded.
Fix $a$ and let $T(x)=\sum_{k=1}^\infty a_kx_k$. For $n=1,2,\dots$ define linear functionals $T_nx = \sum_{k=1}^n a_kx_k$. These are bounded on $\ell^2$ and satisfy $\sup_n |T_n(x)|<\infty$ for every $x$, because the series $\sum_{k=1}^\infty a_kx_k$ converges. By the uniform boundedness principle, also known as the Banach-Steinhaus theorem, we have $\sup_n\|T_n\|<\infty$. Since $T_n\to T$ pointwise, it follows that $\|T\|\le \sup_n\|T_n\|$, and the conclusion follows.
For completeness, I sketch an alternative approach which uses nothing but calculus. Suppose $\sum_{k=1}^\infty |a_k|^2=\infty$. Then we can choose a sequence $(k_j)$ such that
$$\sum_{k_j\le k<k_{j+1}}^\infty |a_k|^2\ge 2^j$$
for every $j$. Within each block of indices $k_j\le k<k_{j+1}$ define $x_k$ (how?) so that
$$\sum_{k_j\le k<k_{j+1}}^\infty |x_k|^2\le 2^{-j},\qquad \sum_{k_j\le k<k_{j+1}}^\infty a_kx_k = 1$$
Then $x\in \ell^2$ and $\sum_{k=1}^\infty a_kx_k$ diverges.
