Increase input current in series resonant circuit
Well, the RMS of the input current is given by:
$$\overline{\text{I}}_\text{in}=\frac{1}{\sqrt{2}}\cdot\frac{\text{amplitude input voltage}}{\left|\text{input impedance}\right|}=\frac{1}{\sqrt{2}}\cdot\frac{\frac{29.2}{2}}{\left|\underline{\text{Z}}_\text{in}\right|}\tag1$$
Where:
$$\underline{\text{Z}}_\text{in}=1+\frac{1}{\text{j}\cdot2\pi\cdot125\cdot1000\cdot853\cdot10^{-9}}+\text{j}\cdot2\pi\cdot125\cdot1000\cdot1.9\cdot10^{-3}+$$
$$4+\frac{1}{\text{j}\cdot2\pi\cdot125\cdot1000\cdot1.39\cdot10^{-9}}=5+\left(475\pi-\frac{341756000}{118567 \pi }\right)\text{j}\tag2$$
So, we get:
$$\overline{\text{I}}_\text{in}=\frac{73}{5 \sqrt{2 \left(\left(\frac{341756000}{118567 \pi }-475 \pi \right)^2+25\right)}}\approx0.0179611\space\text{A}\tag3$$

Finding the point where resonance occurs, we get:
$$\hat{\text{f}}=\frac{1}{2\pi}\sqrt{\frac{\text{C}_\text{res}+\text{C}_\text{s}}{\text{C}_\text{res}\text{C}_\text{s}\text{L}}}\tag4$$
I found that using Mathematica:
In[1]:=x = (1/Sqrt[
      2])*(((U)/
       2))/(R1 + (1/(I*2*Pi*f*C1)) + ((1/(I*2*Pi*f*C2)) + 
        R2 + (I*2*Pi*f*L)));
y = FullSimplify[
  Sqrt[(ComplexExpand[Re[x]])^2 + (ComplexExpand[Im[x]])^2], 
  Assumptions -&gt; 
   U &gt; 0 &amp;&amp; R1 &gt; 0 &amp;&amp; R2 &gt; 0 &amp;&amp; C1 &gt; 0 &amp;&amp; C2 &gt; 0 &amp;&amp; L &gt; 0 &amp;&amp; 
    f &gt; 0]; FullSimplify[
 Solve[{D[y, f] == 0, 
   U &gt; 0 &amp;&amp; R1 &gt; 0 &amp;&amp; R2 &gt; 0 &amp;&amp; C1 &gt; 0 &amp;&amp; C2 &gt; 0 &amp;&amp; L &gt; 0 &amp;&amp; f &gt; 0}, 
  f]]

Out[1]={{f -&gt; ConditionalExpression[Sqrt[(C1 + C2)/(C1 C2 L)]/(2 \[Pi]), 
    R2 &gt; 0 &amp;&amp; R1 &gt; 0 &amp;&amp; C2 &gt; 0 &amp;&amp; C1 &gt; 0 &amp;&amp; L &gt; 0 &amp;&amp; U &gt; 0]}}

At that point we get:
$$\overline{\text{I}}_\text{in}\left(\hat{\text{f}}\right)=\frac{\text{V}_\text{in}}{2\sqrt{2}\left(\text{R}_1+\text{R}_2\right)}\tag5$$
Where \$\text{V}_\text{in}\$ is the amplitude of the input voltage and \$\text{R}_1\$ is the first resistor seen from the left side and \$\text{R}_2\$ is the second resistor.
Using your values we get:

$$\hat{\text{f}}=\frac{500000 \sqrt{\frac{854390}{2252773}}}{\pi }\approx98014.3\space\text{Hz}\tag6$$
$$\overline{\text{I}}_\text{in}\left(\hat{\text{f}}\right)=\frac{73}{25 \sqrt{2}}=2.06475\space\text{A}\tag7$$