The limiting form you wrote is a constant with respect to th; hence, c0=ser1 makes sense.
Also, the limiting form is the limit of F as th goes to 1 from the left. F is actually discontinuous at 1.
If you plot your function for an arbitrary value of t you can see it. The limit of F as th goes to 1 from the left converges, but from the right, diverges.
In[]:=
F[th_]:=(Sqrt[t]*(1-2*th-5*th^2+2*th^3+2*(2+t)*th^4+(th*(3+2*th-(3+2*t)*th^2-2*th^3))/E^t))/(4*Sqrt[Pi]*(-1+th)^4*(1+th)^2)+(th*(4-(1+8*t)*th-4*(2+t)*th^2+(2+8*t+4*t^2)*th^3+4*(1+t)*th^4-th^5)*(-DawsonF[Sqrt[-(t/(-1+th^2))]]+DawsonF[th*Sqrt[-(t/(-1+th^2))]]/E^t))/(4*Sqrt[Pi]*(-1+th)^4*(1+th)^2*Sqrt[1-th^2]);
In[]:=
p1=Plot[F[th]/.t1/2,{th,0,1}]
Out[]=
0.2
0.4
0.6
0.8
1.0
0.02
0.04
0.06
0.08
0.10
0.12
In[]:=
p2=Plot[F[th]/.t1/2,{th,1+0.001,2}]
Out[]=
1.2
1.4
1.6
1.8
2.0
-100
-50
50
100
150
200
For several values of t, as well:
In[]:=
p3=Plot3D[F[th],{th,0,2},{t,0,3}]
Out[]=