This question is unclear. Are we supposed to sum the series (if it converges) or merely test for convergence? I will assume the former for the moment and see what we can do with the series to simplify it or reduce it in some way.
Consider the identities tan(A+B)≡(tanA+tanB)/(1-tanAtanB) and tan(A-B)≡(tanA-tanB)/(1+tanAtanB).
Take pairs of terms in the given series: arctan(n+1)-arctan(n).
Let A=arctan(n+1), B=arctan(n). Therefore, tanA=n+1, tanB=n.
tan(A-B)≡(tanA-tanB)/(1+tanAtanB),
so arctan(tan(A-B))=arctan[(tanA-tanB)/(1+tanAtanB)].
arctan(tan(A-B))=A-B=arctan[1/(1+n(n+1))].
In this way each pair of terms reduces to a single term in a new series.
Take the first 4 terms of the given series (call it S₀) as an example:
arctan(2)-arctan(1)+arctan(4)-arctan(3)+... reduces to
arctan(1/3)+arctan(1/13)+..., the first 2 terms of the new series (S₁).
S₁=∑[n∈[1,∞)]arctan(1/(4n²-2n+1))
In a similar way, we can take a pair of terms in S₁ to create another series (S₂).
Let tanA=1/3, tanB=1/13, then:
arctan((1/3)+(1/13))=arctan[((1/3)+(1/13))/(1-(1/3)(1/13))]=
arctan(16/38)=arctan(8/19). This single term is the combination of the first 4 terms of S₀.
S₂=∑[n∈[1,∞)]arctan(4(4n²-3n+1)/(128n⁴-192n³+104n²-24n+3))
We can call S₁ the first generation of S₀, then Sᵣ is the rth generation of S₀. Sᵣ terms combine groups of 2ʳ terms of S₀. As r gets larger the first term is a better approximation of the sum to infinity of the terms in S₀, the given series.
However, the series may not converge, and because S₁ consists of the sum of positive terms, we could apply a convergence test (Maclaurin-Cauchy, for example).
The series appears to converge to about 29 degrees or 0.506 radians, but as n gets larger the ratio of consecutive terms of S₁ approaches 1, indicating possible divergence (inconclusive).
More to follow...