I think you miss the point. Surely, for a suitably chosen value of x, you can obtain a reasonably accurate approximation to the value of HypergeometricPFQ, either by using the MATHEMATICA built-in procedure, or by the Taylor series around x=0, or by the asymptotic expansion around x=Infinity. (In fact, the Taylor series is convergent for any x in this case, and I suspect MATHEMATICA uses it for calculating HypergeometricPFQ, although this is certainly inefficient for large x, which manifests itself by numerical difficulties and increasing computational times for large x). However, I am trying to draw your attention to a different fact, namely that response 3 of Wolfram Alpha is NOT the asymptotic expansion for which I asked (by calling Series with {x,Infinity,5}, although it is returned as such. This is a mistake which should be corrected. Whether or not the three methods/formulae yield nearly the same numerical result does not matter. Leslaw.

One more comment: When someone enters a command to obtain a series expansion around Infinity, then it should be obvious that one wants to obtain an asymptotic expansion, not a Taylor series. In such a situation, the returned answer 1: "no series expansion available" is simply misleading and should also not be displayed by Wolfram Alpha. It is also inconsistent with answer 2. Leslaw

Did you test this? At all? (These are rhetorical questions.)

Here are the relevant computations.

ee = HypergeometricPFQ[{1, 1, 1, 1}, {2, 2, 2, 2}, -x];

First the series from the second pod.

ss = Normal[Series[ee, {x, Infinity, 5}]]

(* Out[477]= 
E^-x (-(6/x^5) + 1/x^4) + (1/(
 12 x))(2 EulerGamma^3 + EulerGamma \[Pi]^2 + 
   6 EulerGamma^2 Log[x] + \[Pi]^2 Log[x] + 6 EulerGamma Log[x]^2 + 
   2 Log[x]^3 - 2 PolyGamma[2, 1]) *)

Quick numerical check.

{ee, ss} /. x -> 1111.1

(* Out[479]= {0.0715749986818, 0.0715749986818} *)

Good so far. Now for that third pod. I rewrote ever so slightly to use the more common positive exponent notation.

ss2 =  Sum[(-1)^n*x^n/((n + 1)^4*Factorial[n]), {n, 0, Infinity}]
ss2 === ee

(* Out[480]= HypergeometricPFQ[{1, 1, 1, 1}, {2, 2, 2, 2}, -x]

Out[481]= True *)

To restate, it automatically evaluated to the input expression. We can check numerically that this is plausible.

terms = 4000;
NSum[(-1)^n*(1111 + 1/10)^n/((n + 1)^4*Factorial[n]), {n, 0, terms}, 
 WorkingPrecision -> 1000, NSumTerms -> terms]

(* Out[504]= 
0.07157499868179492615228113024556896152461872165957332195911424357595\
7027394013518883378928089038297513792617279360234433705402207531750966\
0717006728987049036451842823432573305827850163550084804675769957660934\
4824128631596694605973285589988236274789694902987106035520518673838341\
8403680202985945005798435970592342850033187925346634452756821088978508\
5259444658718487981866392080437473574473862047376793571126464969800787\
8803181449171740731887666924329532184773807981764974033314775480749947\
1096391441816719897059492637173806116477129407659923386426954912528103\
9169790296933292120323513441841312866076886262356996247624553397188136\
7527417883237598541289814556372772783972646774583249135592744712362526\
9777934478025389513686897128733758151270977948985716655740364364858294\
3580719151581187700654582635529859694967679171786536858445379972974013\
6089392197806995711217621078952538615859473022861414912694258074963677\
5180352839003018355521752110323676919920539509259744224721471269246962\
90624132618936092170922 *)

(Exercise: Show that 4000 terms suffices to get a good approximation.)

Or use Sum with a numeric input for x:

Sum[(-1)^n*(1111.1)^n/((n + 1)^4*Factorial[n]), {n, 0, Infinity}]

(* Out[495]= 0.0715749986818 *)

Since careful verification is the order of the day, I'll show the variant from the W|A third pod as well:

Sum[(-1)^(-n)*x^(-n)/((n - 1)^4*Factorial[-n]), {n,  0, -Infinity, -1}]

(* Out[505]= HypergeometricPFQ[{1, 1, 1, 1}, {2, 2, 2, 2}, -x] *)

Regarding the nature of series expansions: many functions have Puiseux expansions at infinity. This one does not. I agree the first pod either should not be present, or should state more clearly what is intended. But that does not make it wrong, just a bit confusing when taken in tandem with the next two.