Since you are using Mathematica, you definitely want to take a look at the extremely useful package [HolonomicFunctions][1] by Christoph Koutschan.

In your particular example,

    Annihilator[f1[d], {S[d]}]

shows that this function satisfies the recurrence
\begin{equation}
  (4+4d)f_1(d+4)+(4-7d)f_1(d+2)-(6-3d)f_1(d)=0.
\end{equation}

Once known, Mathematica itself can check symbolically that both of your functions satisfy this recurrence:

    (4+4d)f1[d+4] + (4-7d)f1[d+2] - (6-3d)f1[d] // FullSimplify
    (4+4d)f2[d+4] + (4-7d)f2[d+2] - (6-3d)f2[d] // FullSimplify

After checking initial conditions (which Mathematica can do) it follows that $f_1(d)=f_2(d)$ for all even integers $d$

But as I'm typing I see that the OP just figured all of this out by himself... ;)  So let me just mention that one strategy now could be to look at $f_1-f_2$, show that it satisfies the necessary exponential growth conditions (should be alright after combining the poles; apart from these each function seems to be good by itself), and apply [Carlson's Theorem][2]. I hope that helps...


  [1]: http://www.risc.jku.at/research/combinat/software/ergosum/RISC/HolonomicFunctions.html
  [2]: http://en.wikipedia.org/wiki/Carlson%27s_theorem