Electrical – Analyzing LC circuit with multiple inductors and capacitors

KCL: (\$v_o\$ over your output, \$v_i\$ over voltage source and \$v\$ over the internal node:

$$ \frac{v_o}{R_3} + \frac{v_o-v}{Z_{C1}}+\frac{v_o-v_i}{R_2} = 0 \\ \frac{v}{Z_L} + \frac{v-v_i}{Z_{C2}} + \frac{v-v_o}{Z_{C1}} = 0 $$

can be solved to get rid of the internal node \$v\$, setting \$Z_{C i} = \frac{1}{sC_i}\$ and \$Z_L = sL\$:

$$ \frac{v_o}{v_i} = \frac{L C_1 C_2 R_2 R_3 s^3 + L (C_1 + C_2) R_3 s^2+R_3}{ L C_1 C_2 R_2 R_3 s^3 + L(R_2+R_3)(C_1+C_2) s^2 + (R_2 R_3 C_1) s + (R_2 + R_3)} $$ which is in the form $$ \frac{s^3+n_2s^2+n_0}{s^3+d_2 s^2 + d_1 s + d_0}$$ with $$ n_2 = \frac{(C_1+C_2)}{C_1 C_2 R_2} = \frac{1}{R_2C_*}$$ $$ n_0 = \frac{1}{LC_1 C_2 R_2} $$ $$ d_2 = \frac{(R_2+R_3)(C_1+C_2)}{C_1 C_2 R_2 R_3} = \frac{1}{R_\| C_*}$$ $$ d_1 = \frac{1}{LC_2} $$ $$ d_0 = \frac{R_2+R_3}{LC_1 C_2 R_2 R_3} =\frac{1}{LC_1 C_2 R_\|}$$ where I introduced the parallel resistance and serial capacitances $$ \frac{1}{R_\|} = \frac{1}{R_2} + \frac{1}{R_3}\\ \frac{1}{C_*} = \frac{1}{C_1} + \frac{1}{C_2} $$

This is a third-order high-pass filter, so general second order LRC intuition does not apply. You have to plug in your values and show them on a Bode-plot in order to find oscillation problems.

Appendix

Maxima script:

sol  : solve([v_o/R3+(v_o-v)*s*C1+(v_o-v_i)/R2=0, 
              v/(s*L)+(v-v_i)*s*C2 + (v-v_o)*s*C1 = 0], [v_i, v_o]) $
trf  : ev(v_o / v_i, sol) $
res  : rat(trf, s);

What effect will it have on my resonant frequency

Theoretically it will have no effect on the resonant frequency. The resonant frequency is purely determined by the capacitor having exactly the opposite reactance of the inductor at a particular frequency and the two reactances cancel leaving the series tuned circuit having only resistance at resonance.

However, with lower values of resistance the peak shape of the resonance will change but the centre point of the peak will remain as previous.