Computing a series in terms of exponential function

Is there any way to compute the following series in terms of exponential function ? $$\sum_{k=0}^\infty Y_1(k)\;x^k$$ where

$$Y_1(k) = \frac{(k - 1)!}{k!}Y_3(k - 1)$$ $$Y_2(k) = \frac{(k - 1)!}{k!}Y_4(k - 1)$$ $$Y_3(k) = \frac{(k - 1)!}{k!}(1/2Y_3(k - 1) + 1/2Y_1(k - 1))$$ $$Y_4(k) = \frac{(k - 1)!}{k!}(1/2Y_4(k - 1) + 1/2Y_2(k - 1))$$ such that: $\;Y_1(0)= 2 + r,\; Y_2(0) = 4 - r,\; Y_3(0) = 2 + r,\; Y_4(0) = 4 - r$

Y1[0] := 2 + r;
Y2[0] := 4 - r;
Y3[0] := 2 + r;
Y4[0] := 4 - r;
Y1[k_] := (k - 1)!/(k!)*(Y3[k - 1]);
Y2[k_] := (k - 1)!/(k!)*(Y4[k - 1]);
Y3[k_] := (k - 1)!/(k!)*(1/2*Y3[k - 1] + 1/2*Y1[k - 1]);
Y4[k_] := (k - 1)!/(k!)*(1/2*Y4[k - 1] + 1/2*Y2[k - 1]);

Sum[ Y1[k] x^k, {k, 0, Infinity}]

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Your coefficients aren't monotonic ListLinePlot@ Table[Table[FullSimplify[Y1[k]], {k, 0, 10}] /. r -> j, {j, 6}] – Dr. belisarius Nov 21 '13 at 13:23
@belisarius, This series is convergent to $(2+r)e^x$. Why can not Mathematica compute it? – Angel Nov 22 '13 at 8:24
Are you sure? i.stack.imgur.com/V1yCd.png – Dr. belisarius Nov 22 '13 at 12:04

Edit

You changed your equations after accepting this answer. The same method works:

First we modify your functions to get some speed:

Y1[0] = 2 + r;
Y2[0] = 4 - r;
Y3[0] = 2 + r;
Y4[0] = 4 - r;
Y1[k_] := Y1[k] = FullSimplify[(k - 1)!/(k!)*(Y3[k - 1])];
Y2[k_] := Y2[k] = FullSimplify[(k - 1)!/(k!)*(Y4[k - 1])];
Y3[k_] := Y3[k] = FullSimplify[(k - 1)!/(k!)*(1/2*Y3[k - 1] + 1/2*Y1[k - 1])];
Y4[k_] := Y4[k] = FullSimplify[(k - 1)!/(k!)*(1/2*Y4[k - 1] + 1/2*Y2[k - 1])];


Now we calculate the coefficients (we take out a (2+r) factor):

f = FindSequenceFunction[Table[FullSimplify[Y1[i]], {i, 0, 50}], n]
(*
(2 + r)/Pochhammer[1, -1 + n]
*)

Sum[f x^n, {n, 0, Infinity}]
(*
E^x (2 + r) x
*)

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Your codes don't work. – Angel Nov 23 '13 at 19:28
@Angel Check it now. There was a syntax error. – Dr. belisarius Nov 23 '13 at 20:23
Excellent. I edited my question.Thanks a million. – Angel Nov 25 '13 at 14:34