The discussion focuses on evaluating the sum of the first n terms of the series $\displaystyle \sum_{i=0}^n \tan^{-1} \dfrac{1}{i^2+i+1}$. A notable solution presented by MarkFL involves a generalization of the series, expressed as $$ \sum_{k=0}^{n} \arctan \left( \frac{a}{a^{2}k^{2}+a(a+2b)k +(1+ab+b^{2})} \right) = \arctan \Big(a(n+1)+b \Big) - \arctan(b)$$. This formula applies under the condition that $ak+b >0$, providing a structured approach to solving similar arctangent summation problems.
PREREQUISITESMathematicians, students studying calculus, and anyone interested in advanced summation techniques and series analysis.
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 $ak+b >0$