I am trying to show that the linear system $H x = b$, where $H$ is a Hilbert matrix of size $n \times n$ and
$$ b_{i} = \sum_{j=1}^{n} \frac{1}{i+j-1} $$
has the solution $x = (1,1,\dots,1)$.
Looking at the multiplication $H x$, I don't see how $(1,1,\dots,1)$ could be the solution, since after doing the second row in the matrix multiplication $1/2 + 1/3 + 1/4 + \cdots$ would be greater than the second entry in $b_{i}$, which is $1/2$.
You're misreading the definition of $b_i$. Each $b_i$ is a sum of $n$ terms; $b_2=\frac12+\frac13+\frac14+\cdots+\frac1{n+1}$, for example. There are two indices there - the $i$ to mark which entry of $b$ it is, and the $j$ to sum over. The vector $b$ can also be written $$b=\begin{pmatrix}\frac11+\frac12+\frac13+\cdots+\frac1n\\ \frac12+\frac13+\frac14+\cdots+\frac1{n+1}\\ \frac13+\frac14+\frac15+\cdots+\frac1{n+2}\\ \vdots\\ \frac1n+\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n-1}\end{pmatrix}$$ without the $\sum$ notation.