Let $H$ be a separable Hilbert space, and $(\chi_j)_j$ be a sequence in $H$ such that the following operator is a well defined bounded operator:
Consider $T: H \to l^2: f \mapsto (\langle f, \chi_j\rangle)_j$.
We may also assume that $T$ is bijective on its image.
My book then says that the adjunct operator is given by
$$T^*: l^2 \to H: (x_j)_j \mapsto \sum_{j}^\infty x_j \chi_j$$ but I can't see why the series converges.
Note that it is not given that the $(\chi_j)_j$ are orthogonal.
Note if you assume just that $T$ is well defined it follows in any number of ways that $T$ is bounded.
Anyway, that series converges because $T^*$ is continuous. Given $x\in\ell^2$ define $$s_n=(x_1,\dots,x_n,0,0,0,\dots).$$You verify from the definition that $$T^*s_n=\sum_{j=1}^nx_j\chi_j.$$And now since $s_n\to x$ it follows that $T^*x=\lim T^*s_n$.