Evaluate $\sum_{n=1}^{\infty} \left( \frac{\pi}{6} - \int_{0}^{n} \frac{\sqrt{3} (2x+1)}{(x^2+x+3)^2+3} dx \right)$
You may avoid the evaluation of the arctan series. Note that $$f(x) = \frac{\sqrt{3}(2x+1)}{(x^2+x+3)^2+3}=g(x)-g(x+1)$$ where $$g(x)=\frac{\sqrt{3}}{x^2+3}.$$ Moreover $$G(x)=\int_0^x g(t)\,dt=\arctan\left(\frac{x}{\sqrt{3}}\right).$$ Hence, since $G(1)=\frac{\pi}{6}$, it follows that $$\begin{align} \sum_{n=1}^{\infty} \left( \frac{\pi}{6} - \int_{0}^{n} f(x)\, dx \right)&= \sum_{n=1}^{\infty}(G(1)- \int_{0}^{n} (g(x)-g(x+1)) dx) \\ &=\sum_{n=1}^{\infty} (G(1)-(G(n)-(G(n+1)-G(1)))\\ &= \sum_{n=1}^{\infty}(G(n+1)-G(n))\\ &=\lim_{n\to +\infty}G(n)-G(1)=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}.\end{align}$$
I heard the solution from another person. So I post it here.
- It is equivalent to compute $\displaystyle \sum_{n=1}^{\infty} \arctan \left( {\sqrt{3} \over n^2+n+3} \right)$
- Use identity: $$ \arctan \frac{n}{n+4} \sqrt{3} - \arctan \frac{n-1}{n+3}\sqrt{3} = \arctan \frac{\sqrt{3}}{n^2+n+3} $$ to prove that it converges to $\arctan( \sqrt{3}) = \pi/3$.
As you mentioned, the sum is equivalent to $$\sum_{i=1}^\infty \arctan\frac{\sqrt{3}}{n^2+n+3}$$ because $\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}$.
To compute series containing $\arctan$ terms, usually we use the identity $$\arctan x-\arctan y=\arctan\frac{x-y}{1+xy}$$ to turn the sum in telescopic form.
Specially, we look for a sequence $p_n$ such that (in this case): $$\arctan\frac{\sqrt{3}}{n^2+n+3}=\arctan p_n-\arctan p_{n-1}=\arctan \frac{p_n-p_{n-1}}{1+p_np_{n-1}}\qquad(*) $$ and in this case we need to consider $p_n$ in form $\frac{an+b}{cn+d}$ and look for coefficients $a$, $b$, $c$ and $d$ to satisfy $(*)$.