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How can we show that, using either of the two equivalent functions, we converge to some number? $$\sum_{n=0}^\infty f(n)\ =\ 2.75056...$$ $$ f(n):=\frac{1}{\binom{\frac{1}{8} \left(2 (n+2) n+(-1)^{n+1}+1\right)}{\frac{1}{8} \left(2 n^2+(-1)^n-1\right)}}=\frac{\Gamma \left(\frac{1}{4} \left(2 n+(-1)^{n+1}+5\right)\right) \Gamma \left(\frac{1}{8} \left(2 n^2+(-1)^n+7\right)\right)}{\Gamma \left(\frac{1}{8} \left(2 (n+2) n+(-1)^{n+1}+9\right)\right)} $$ Also, assuming it converges, how do we get the first 120 digits?

f[n_] := (Gamma[1/4 (5 + (-1)^(1 + n) + 2 n)] Gamma[1/8 (7 + (-1)^n + 2   n^2)])/
  Gamma[1/8 (9 + (-1)^(1 + n) + 2 n (2 + n))]

Sum[f[n],{n,0,Infinity}]

Edit 120-digits:

2.75056595798725797215253771065184065733930718965631096586548358464826\ 860136337395324745249086656444919287115242994885769

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You can get rid of the problematic powers of -1 by combining pairs. Then do a ration test. I only do an approximation below so this rests on Series not dropping important terms e.g. from unrecognized cancellation. My guess is it's fine, and some numeric tests indicate the approximation below is reliable. It shows that ratios of paired sums decreases as O(1/n), so this converges, and does so geometrically.

f1[n_] := (Gamma[1/4 (5 - 1 + 2 n)] Gamma[1/8 (7 + 1 + 2 n^2)])/
  Gamma[1/8 (9 - 1 + 2 n (2 + n))]
f2[n_] := (Gamma[1/4 (5 + 1 + 2 n)] Gamma[1/8 (7 - 1 + 2 n^2)])/
  Gamma[1/8 (9 + 1 + 2 n (2 + n))]
ratio = (f1[n] + f2[n + 1])/(f1[n + 2] + f2[n + 3]);
ratioApprox = Normal[Series[ratio, {n, Infinity, 1}]];
r2 = ExpandAll[ratioApprox]

(* Out[569]= -(5/12) E^(
  2 - 32944/(21 n^5) + 1064/(3 n^4) - 422/(5 n^3) + 64/(3 n^2) - 6/
   n) + E^(1 - 32944/(21 n^5) + 1064/(3 n^4) - 72/n^3 + 50/(3 n^2) - 
  4/n) + 1/2 E^(
  2 - 32944/(21 n^5) + 1064/(3 n^4) - 422/(5 n^3) + 64/(3 n^2) - 6/n)
   n *)

From this one can also deduce a bound on how many terms are needed to get 120 digits.

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I can only answer to the first part of your question. One can show that the function converge to a number

1.) with a plot (first guess)

f = 1/Binomial[1/8 (2 n (n + 2) + (-1)^(n + 1) + 1), 1/8 (2 n^2 + (-1)^n - 1)]
DiscretePlot[Sum[f, {n, 0, k}], {k, 25}]

enter image description here

2.) with EulerSum in the NumericalCalculus package. Unfortunately works EulerSumonly with MachinePrecision.

<< NumericalCalculus`
EulerSum[f[n], {n, 0, \[Infinity]}] // NumberForm[#, 16] &
(* 2.750565024528173 *)
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