Trigonometric Summation!

Solution :

$n^{2} + n + 1 = n \cdot (n + 1) + 1$

Therefore:

Using the identity : $arctan(x) - arctan(y) = arctan(\frac{x - y}{1 + x \cdot y})$ ;

Let $x = n + 1$ and $y = n$ .

Then,

$arctan(n + 1) - arctan(n) = arctan(\frac{(n + 1) - n}{n \cdot (n + 1) + 1}) = arccot(\frac{n \cdot (n + 1) + 1}{(n + 1) - n})$ $[arctan(x) = arccot(\frac{1}{x})]$

Thus,

$arccot(n^{2} + n + 1) = arctan(n + 1) - arctan(n)$

So,

$\sum_{n = 1}^{∞} arccot(n^{2} + n + 1) = \sum_{n = 1}^{∞} (arctan(n + 1) - arctan(n))$ ,

The sum is now equal to : $arctan(2) - arctan(1) + arctan(3) - arctan(2) + arctan(4) - arctan(3) +$ .......

This sum telescopes to : $(\lim_{n\to ∞} arctan(n)) - arctan(1)$ , which is equal to $\frac{π}{2} - \frac{π}{4}$ which is equal to $\frac{π}{4}$ .

Solution similar to @Torus Wheel 's

$\begin{aligned} S & = \sum_{n=1}^\infty \text{arccot }(n^2+n+1) \\ & = \sum_{n=1}^\infty \arctan \left(\frac 1{n^2+n+1}\right) \\ & = \sum_{n=1}^\infty \arctan \left(\frac {(n+1)-n}{1+(n+1)n}\right) \\ & = \sum_{n=1}^\infty \left(\arctan (n+1) - \arctan (n)\right) \\ & = \lim_{n \to \infty} \arctan (n+1) - \arctan (1) \\ & = \frac \pi 2 - \frac \pi 4 = \frac \pi 4 \approx \boxed{0.785} \end{aligned}$