What is the limit of the series with n=1 to ∞ Σ1/n^4+4

To obtain the closed form for this series, one can use the Residue Theorem as follows:
Σ(-∞ to ∞) f(n) = - [sum of the residues of all poles of π cot(πz) f(z) at all poles of f(z)].

(This is usually proved in any standard Complex Analysis textbook.)
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Here, f(z) = 1/(z^4 + 4), which only has four simple poles at z = 4^(1/4) e^(πi(2k+1)/4) for k = 0,1,2,3, and is analytic elsewhere. Let's label these by k: z₀, z₁, z₂, z₃.

One can check that these poles reduce to z = ±1 ± i (all four possibilities).

Residues of π cot(πz)/(z^4 + 4) at z = z(k):
lim(z→ z(k)) (z - z(k)) * π cot(πz)/(z^4 + 4)
= lim(z→ z(k)) π cot(πz) * lim(z→ z(k)) (z - z(k))/(z^4 + 4)
= π cot(π z(k)) * lim(z→ z(k)) 1/(4z^3)
= π cot(π z(k)) * lim(z→ z(k)) z/(4z^4)
= π cot(π z(k)) * z(k) / (4 * (-4)), since [z(k)]^4 = -4
= (-π/16) z(k) cot(π z(k)).

Therefore,
Σ(-∞ to ∞) 1/(n^4 + 4)
= - Σ(k = 0 to 3) [(-π/16) z(k) cot(π z(k))
= (π/16) Σ(k = 0 to 3) z(k) cot(π z(k))
= (π/16) [(1+i) cot(π(1+i)) + (-1-i) cot(π(-1-i)) + (1-i) cot(π(1-i)) + (-1+i) cot(π(-1+i))]
= (π/16) [(1+i) cot(π(1+i)) + (1+i) cot(π(1+i)) + (1-i) cot(π(1-i)) + (1-i) cot(π(1-i))]
since cotangent is odd
= (π/8) [(1 + i) cot(π + πi) + (1 - i) cot(π - πi)].
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Simplifying this further:
cot(π ± πi)
= cos(π ± πi) / sin(π ± πi)
= [cos π cos (±πi) - sin π sin(±πi)] / [sin π cos (±πi) + cos π sin(±πi)]
= [-cos (±πi) - 0] / [0 + (-1) sin(±πi)]
= cos (±πi) / sin(±πi)
= cos (πi) / (-sin(πi))
= cosh π / [(1/i) sinh π]
= i coth π.
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So, Σ(-∞ to ∞) 1/(n^4 + 4)
= (π/8) [(1 + i) * i coth π + (1 - i) * i coth π].
= (πi/4) coth π

However, Σ(-∞ to ∞) 1/(n^4 + 4)
= Σ(-∞ to -1) 1/(n^4 + 4) + 1/(0^4 + 4) + Σ(1 to ∞) 1/(n^4 + 4)
= Σ(1 to ∞) 1/(n^4 + 4) + 1/4 + Σ(1 to ∞) 1/(n^4 + 4), via evenness
= 2 * Σ(n = 1 to ∞) 1/(n^4 + 4) + 1/4.

Hence, 2 * Σ(n = 1 to ∞) 1/(n^4 + 4) + 1/4 = (πi/4) coth π