Create variables in a for loop

Clear["Global`*"]

Continuing on with the suggestion by infinitezero to use recursion, use RSolve to find the general expression.

ex[n_] = ex[n] /. 
  RSolve[{ex[0] == 1, ex[n] == ex[n - 1] + (1/2)^n/n!}, ex[n], n][[1]]

(* (Sqrt[E] Gamma[1 + n, 1/2])/Gamma[1 + n] *)

The first several values are

ex /@ Range[0, 10] *)

(* {1, 3/2, 13/8, 79/48, 211/128, 6331/3840, 75973/46080, 354541/215040, \
17017969/10321920, 306323443/185794560, 2042156287/1238630400}

Which are approximately,

% // N

(* {1., 1.5, 1.625, 1.64583, 1.64844, 1.6487, 1.64872, 1.64872, 1.64872, \
1.64872, 1.64872} *)

%[[-1]] // InputForm

(* 1.6487212706873657 *)

The limit of this sequence is

Limit[ex[n], n -> Infinity]

(* Sqrt[E] *)

% // N[%, 20] &

(* 1.6487212707001281468 *)

What about defining this recursively?

ex[0] = 1;
ex[n_] := ex[n] = ex[n - 1] + 0.5^n/n!;

The double assignment in the second line is very important. This causes all function calls to be evaluated only once. Once it has been initially evaluated, it will be saved in ex[n].

Table[{k, ex[k]}, {k, 1, 10}] // TableForm

1   1.5
2   1.625
3   1.64583
4   1.64844
5   1.6487
6   1.64872
7   1.64872
8   1.64872
9   1.64872
10  1.64872

One more way, using FoldList:

FoldList[#1 + 0.5^#2/#2! &, 1 , Range[7]]
(* {1, 1.5, 1.625, 1.64583, 1.64844, 1.6487, 1.64872, 1.64872} *)