As stated in the title I am trying to prove $g_{n}=\sum_{k=1}^{2^n}\frac{2^{n}}{k}\chi_{\left(\left(\frac{k-1}{2^{n}}\right)^{2},\left(\frac{k}{2^{n}}\right)^{2}\right]}$ converges pointwise to $\frac{1}{\sqrt{x}}$ on $(0,1]$. Obviously $\left\lbrace\left(\left(\frac{k-1}{2^{n}}\right)^{2},\left(\frac{k}{2^{n}}\right)^{2}\right]\right\rbrace_{k=1}^{2^n}$ is a partition of the interval $(0,1)$, so for some $k\in\left\lbrace 1,2,\dots,2^n\right\rbrace$, $g_n(x) =\frac{2^n}{k}$.
I was thinking about taking the largest $n$ such that for some $k\in\left\lbrace 1,2,\dots,2^n\right\rbrace$, $x\in \left(\left(\dfrac{k-1}{2^n} \right)^2,\left(\dfrac{k}{2^n}\right)^2 \right)$ and then breaking that up to find a $N$ such that $x=\left(\dfrac{k}{2^N}\right)^2$, but I am not sure such a $n$ exists. Any help is welcome. I tried squeeze theorem as well, but I failed to find one of the bounds.
For any $n$ and $x$ there is a unique $k_n$ such that $(\frac {k_n-1} {2^{n}})^{2} <x \leq (\frac {k_n} {2^{n}})^{2}$. Note that $\frac {k_n-1} {2^{n}} <\sqrt x \leq \frac {k_n} {2^{n}}$. This implies that $\frac {k_n} {2^{n}} \to \sqrt x$. Hence $g_n(x)=\frac {2^{n}} {k_n} \to \frac 1 {\sqrt x}$.