Well, 1st in your expression there are square-brackets around Epsilon. That may give difficulties beacuse of the special use of square-brackets in Mathematica. If you delete them you get

In[2]:= Hypergeometric1F1[
 1.99997 - (0.00729735 Epsilon)/Sqrt[1 - Epsilon^2], 2.99995, 
 274.449 Sqrt[1 - Epsilon^2]]


Out[2]= Hypergeometric1F1[
 1.99997 - (0.00729735 Epsilon)/Sqrt[1 - Epsilon^2], 2.99995, 
 274.449 Sqrt[1 - Epsilon^2]]

So as remarked by Sean Epsilon is not defined. If you do that you get a (quite big) number

 In[4]:= Epsilon = .01; 
Hypergeometric1F1[  1.99997 - (0.00729735 Epsilon)/Sqrt[1 - Epsilon^2], 2.99995, 
  274.449 Sqrt[1 - Epsilon^2]]


 Out[4]= 1.11327*10^117

Regards, Hans

Functions like Hypergeometric1F1 will evaluate with variables in their arguments if there's a simple transformation.

1. Is that the case here? What are you expecting it to return? Are you sure there's a simple answer?

2. Very often, these simplifications are symbolic and require that you not use floating point numbers. You have used floating point numbers. So it probably won't do any symbolic simplification. Symbolic transformations generally aren't safe to perform on finite precision numbers. You can convert from floating point to rational numbers (which are symbolic) using rationalize.

3. You may need to use FunctionExpand or FullSimplify to search for your answer. FunctionExpand won't work in this case. FullSimplify typically requires that you list out assumptions with the Assumptions option to work well.

I'm not sure I understand. Do you want to express the Hypergeometric1F1 as an infinite series?

Epsilon isn't defined anywhere. So it can't evaluate further.