The value of the sum of the series ${\sum }_{n=1}^{\infty }\arctan \left(\frac{1}{n^2+n+1}\right)$ is

Solution

1. Use the trigonometric identity $\arctan (x)-\arctan (y)=\arctan \left(\frac{x-y}{1+xy}\right)$ to rewrite the general term.
2. Observe that $\frac{1}{n^2+n+1}=\frac{(n+1)-n}{1+n(n+1)}$, so the term becomes $\arctan (n+1)-\arctan (n)$.
3. The partial sum $S_N={\sum }_{n=1}^N(\arctan (n+1)-\arctan (n))$ simplifies to $\arctan (N+1)-\arctan (1)$ due to the telescoping property.
4. Taking the limit as $N\to \infty$, we get $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$.
5. Hence the answer is (A).