Suppose $P_n$ is a monotone sequences of orthogonal projections in a complex Hilbert space $\mathcal{H}$. I want to show that the limit of the sesquilinear forms defined by:
$\Gamma_n(x,y)=(x,P_n y)$ where $x,y$ are in $\mathcal{H}$.
That is the limit defined by $\Gamma(x,y)= \lim_{n \rightarrow \infty} \Gamma_n(x,y)$ exists.
I think I ought to use the fact that the Hilbert space is complete in some sense and that the inner product is continuous.
Note: There is nothing about completeness of $\mathcal{H}$ needed to carry out of the following steps.
Because $P_n$ is monotone, then $(P_nx,x)$ is monotone in $n$ for each fixed $x$, and is bounded above by $(x,x)$, which forces convergence of $\lim_n(P_n x,x)$ for all $x$. Then, using polarization, the following expression must also have a limit in $n$ for every fixed $x,y \in \mathcal{H}$: $$ (P_nx,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}(P_n(x+i^ny),x+i^n y) $$ Therefore the following limit exists for all $x,y\in\mathcal{H}$: $$ \Gamma(x,y) = \lim_n (P_nx,y). $$ Because the limit exists for all $x,y$, you can show that $\Gamma$ is a sesquilinear form. And it's easy to check that $0 \le \Gamma(x,x)=\overline{\Gamma(x,x)} \le (x,x)$ for all $x$ because this is inherited from $(P_nx,x)$.
If you want to represent $\Gamma(x,y)=(Px,y)$ for some bounded linear operator $P$, then you'll need completeness.