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

Here's the fix:
cot(π ± πi)
= cos (±πi) / sin(±πi)
= cos (πi) / (±sin(πi))
= cosh π / [(∓1/i) sinh π]
= ∓i coth π.

Σ(-∞ to ∞) 1/(n^4 + 4)
= (π/8) [(1 + i) * -i coth π + (1 - i) * i coth π].
= (π/4) coth π

Now that the i is gone,
Σ(n = 1 to ∞) 1/(n^4 + 4) = (1/8) [π coth π - 1].

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It is 4+(1/90)(pi^4)
There is a general result for sum of 1/(n^(2k)) in terms of Bernoulii coefficient B(2k) as
(-1)^(k-1) *(pi^2k)*(2^2k)*B(2k)/[2*(2k!)] and B(4)=1/30

using this sum of 1/n^4 is 1/90 (pi^4)