Since phasors sum up vectorially, I believe it is going to read \$ (V²+(V ᶜ)²+2VV^{c}\cosθ)^{\frac{1}{2}}\$, where \$V\$ is the source maximum voltage, \$ V^{c}\$ is the capacitor maximum voltage and \$ \theta \$ is the phase angle of the source's voltage phasor relative to capacitor's voltage phasor.
Is that correct?
Well, notice that the amplitude of the voltage across the capacitor is given by:
\begin{equation} \begin{split} \hat{\text{V}}_\text{C}&=\left|\underline{\text{V}}_{\space\text{C}}\right|\\ \\ &=\left|\underline{\text{Z}}_{\space\text{C}}\cdot\underline{\text{I}}_{\space\text{C}}\right|\\ \\ &=\left|\underline{\text{Z}}_{\space\text{C}}\right|\cdot\left|\underline{\text{I}}_{\space\text{C}}\right|\\ \\ &=\left|\frac{1}{\text{j}\omega\text{C}}\right|\cdot\left|\underline{\text{I}}_{\space\text{source}}\right|\\ \\ &=\left|\frac{1}{\text{j}\omega\text{C}}\right|\cdot\left|\frac{\underline{\text{V}}_{\space\text{source}}}{\underline{\text{Z}}_{\space\text{i}}}\right|\\ \\ &=\frac{\left|1\right|}{\left|\text{j}\omega\text{C}\right|}\cdot\frac{\left|\underline{\text{V}}_{\space\text{source}}\right|}{\left|\underline{\text{Z}}_{\space\text{i}}\right|}\\ \\ &=\frac{1}{\omega\text{C}}\cdot\frac{\displaystyle\hat{\text{V}}_{\space\text{source}}}{\displaystyle\left|\text{R}+\text{j}\omega\text{L}+\frac{1}{\text{j}\omega\text{C}}\right|}\\ \\ &=\frac{1}{\omega\text{C}}\cdot\frac{\displaystyle\hat{\text{V}}_{\space\text{source}}}{\displaystyle\left|\text{R}+\text{j}\omega\text{L}-\frac{\text{j}}{\omega\text{C}}\right|}\\ \\ &=\frac{1}{\omega\text{C}}\cdot\frac{\displaystyle\hat{\text{V}}_{\space\text{source}}}{\displaystyle\left|\text{R}+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)\text{j}\right|}\\ \\ &=\frac{1}{\omega\text{C}}\cdot\frac{\displaystyle\hat{\text{V}}_{\space\text{source}}}{\displaystyle\sqrt{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)^2}} \end{split}\tag1 \end{equation}
So, the voltage across the source and capacitor is given by:
\begin{equation} \begin{split} \hat{\text{V}}_{\space\text{source + capacitor}}&=\hat{\text{V}}_{\space\text{source}}+\hat{\text{V}}_\text{C}\\ \\ &=\hat{\text{V}}_{\space\text{source}}+\frac{1}{\omega\text{C}}\cdot\frac{\displaystyle\hat{\text{V}}_{\space\text{source}}}{\displaystyle\sqrt{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)^2}}\\ \\ &=\hat{\text{V}}_{\space\text{source}}\left(1+\frac{1}{\omega\text{C}}\cdot\frac{\displaystyle1}{\displaystyle\sqrt{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)^2}}\right) \end{split}\tag2 \end{equation}
EDIT:
Now, we can solve for the maximum:
$$\frac{\displaystyle\partial}{\displaystyle\partial\omega}\left(\hat{\text{V}}_{\space\text{source}}\left(1+\frac{1}{\omega\text{C}}\cdot\frac{\displaystyle1}{\displaystyle\sqrt{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)^2}}\right)\right)=0\tag3$$
Solving for \$\omega\$, gives:
$$\hat{\omega}=\frac{\displaystyle1}{\displaystyle\text{L}}\cdot\sqrt{\frac{\displaystyle\text{L}}{\displaystyle\text{C}}-\frac{\displaystyle\text{R}^2}{2}}\space,\space\text{CR}^2<2\text{L}\tag4$$
So, for the maximum we get:
\begin{equation} \begin{split} \max\hat{\text{V}}_{\space\text{source + capacitor}}&=\lim_{\omega\space\to\space\hat{\omega}}\hat{\text{V}}_{\space\text{source}}\left(1+\frac{1}{\omega\text{C}}\cdot\frac{\displaystyle1}{\displaystyle\sqrt{\text{R}^2+\left(\omega\text{L}-\frac{1}{\omega\text{C}}\right)^2}}\right)\\ \\ &=\hat{\text{V}}_{\space\text{source}}\left(1+\frac{\displaystyle2\text{L}}{\displaystyle\text{R}\sqrt{\text{C}\left(4\text{L}-\text{CR}^2\right)}}\right) \end{split}\tag5 \end{equation}
