The sum ∑n=1∞ tan−1⁡1n2+n+1 is equal to

∑n=1∞ tan−1⁡1n2+n+1=∑n=1∞ tan−1⁡(n+1)−(n)1+(n+1)(n)=∑n=1∞ tan−1 (n+1)-tan-1n=tan-12-tan-11+tan-13-tan-12+.......tan-1∞=π2−tan−1⁡1=π4