transistors - Finding current through an LED in an amplifier using a BJT

Suppose you have the schematic:

My job is to find $I_{LED} $ knowing that the impedance of the capacitor in negligible against $V_s$. Here's my attempt:

I intend to do an AC analysis followed by a DC analysis, and get both components of the current that goes by the LED.

Where $R_A=R_1||R_2$

For the AC analysis: I did the schematic in the figure and with the following set of equations I get to the final result:

$$i_1=i_2+i_B \\v_s-i_1R_B-i_2R_A=0 \\ v_s-i_1R_B-i_ER_E=0 \\ i_C=\beta i_B \\ i_C\approx i_E \\ \therefore i_C=\frac{\frac{v_s}{R_B}}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_B}\bigg)}$$

For the DC analysis I use the following set, reaching the result:

$$V_cc-I_1R_1-I_2R_2=0 \\ V_cc-I_1R_1-V_{BE}-I_ER_E=0 \\ I_1-I_2=I_B\\ I_B=\frac{I_C}{\beta} \\ I_C \approx I_E \\ \therefore I_C=\frac{\frac{V_{cc}}{R_1}-V_{BE}\bigg(\frac{1}{R_1}+\frac{1}{R_2}\bigg)}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_1}+\frac{1}{R_2}\bigg)}$$

Finally I get to:

$$I_{LED}=I_C+i_c=\frac{\frac{V_{cc}}{R_1}-V_{BE}\bigg(\frac{1}{R_1}+\frac{1}{R_2}\bigg)}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_1}+\frac{1}{R_2}\bigg)}+\frac{\frac{v_s}{R_B}}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_B}\bigg)}$$

However the correct solution is:

$$I_{LED}=\frac{\frac{V_{cc}}{R_1}-V_{BE}\bigg(\frac{1}{R_B}+\frac{1}{R_1}+\frac{1}{R_2}\bigg)}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_B}+\frac{1}{R_1}+\frac{1}{R_2}\bigg)}+\frac{\frac{v_s}{R_B}}{\frac{1}{\beta}+R_E\bigg(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_B}\bigg)}$$

I know that this is a very long question and I don't ask for a solution to the problem, because I have one solution that leads to the correct result, however they don't do the DC and AC analysis separately and I was trying to do it because it should work, but I don't know where the mistake is, the answers are really similar and the AC analysis appears to be correct but the DC does not, and I can't find the mistake.

EDIT: I have found the mistake, In the DC analysis schematics I left out the resistance $R_B$, and if I leave it in the schematics connected to the ground then I can substitute, in the DC analysis only, all the $R_2$´s by the paralel resistance of $R_B$ and $R_2$ and the answers match. But a new question comes up, why do I have to leave $R_B$ connected to the ground? I thought that with a DC analisys capacitors would be replaced by an open circuit, therefore, no current should go through $R_B$, and that's why I ignored it. And this is just th case where I have a resistance $R_B$ there, if I didn't, but left that branch connected to the ground, then wouldn't it act as a short and we could ignore $R_2$?