The discussion revolves around evaluating the sum of the series $\displaystyle \sum_{i=0}^n \tan^{-1} \dfrac{1}{i^2+i+1}$. Participants explore different approaches and generalizations related to this summation, including potential formulas and methods for simplification.
There is no clear consensus on the evaluation of the sum or the proposed generalizations, as multiple approaches and expressions are presented without agreement on their validity or applicability.
The discussion includes generalizations that depend on specific conditions, such as $ak+b >0$, which may not be universally applicable without further clarification.
Participants interested in mathematical series, arctangent functions, and generalizations in summation techniques may find this discussion relevant.
MarkFL said:My solution:
We are given to evaluate:
$$S_n=\sum_{k=0}^n\left[\tan^{-1}\left(\frac{1}{k^2+k+1} \right) \right]$$
Using the identity:
$$\tan^{-1}(x)=\cot^{-1}\left(\frac{1}{x} \right)$$
we may write:
$$S_n=\sum_{k=0}^n\left[\cot^{-1}\left(k^2+k+1 \right) \right]$$
Now, using the fact that:
$$k^2+k+1=\frac{k(k+1)+1}{(k+1)-k}$$
and the identity:
$$\cot(\alpha-\beta)=\frac{\cot(\alpha)\cot(\beta)+1}{\cot(\beta)-\cot(\alpha)}$$
We may now write:
$$S_n=\sum_{k=0}^n\left[\cot^{-1}(k)-\cot^{-1}(k+1) \right]$$
This is a telescoping series, hence:
$$S_n=\cot^{-1}(0)-\cot^{-1}(n+1)=\frac{\pi}{2}-\cot^{-1}(n+1)$$
Using the identity:
$$\tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2}$$
We may also write:
$$S_n=\tan^{-1}(n+1)$$
And so we have found:
$$\sum_{k=0}^n\left[\tan^{-1}\left(\frac{1}{k^2+k+1} \right) \right]=\tan^{-1}(n+1)$$
Random Variable said:A generalization is $$ \sum_{k=0}^{n} \arctan \left( \frac{a}{a^{2}k^{2}+a(a+2b)k +(1+ab+b^{2})} \right) = \arctan \Big(a(n+1)+b \Big) - \arctan(b)$$
where $ak+b >0$