How to calculate $\sum_{k=1}^{\infty}\operatorname{arccot}\frac{1-k^2+k^4}{2k}$

Question

How to calculate this trigonometric sum?

$$\sum_{k=1}^{\infty}\operatorname{arccot}\frac{1-k^2+k^4}{2k}$$

Hint: $\frac{2k}{1-k^2+k^4} = \frac{k(k+1)-k(k-1)}{1 + k(k+1)\cdot k(k-1)}$ — achille hui, Feb 21 '16 at 17:56
Excellent hint @achille hui ; in connexion with this hint, I add another one: $cot(a-b)=\dfrac{cot(a)cot(b)-1}{cot(a)+cot(b)}$... — Jean Marie, Feb 21 '16 at 18:37

score 1 · Accepted Answer · answered Mar 09 '17 at 18:36

1

As suggested by Achille Hui, $$ \text{arccot}\frac{1-k^2+k^4}{2k} = \arctan(k(k+1))-\arctan(k(k-1)) $$ hence the given series is a telescopic series in disguise, converging to $\color{red}{\large\frac{\pi}{2}}$.

answered Mar 09 '17 at 18:36

Jack D'Aurizio

353,855

How to calculate $\sum_{k=1}^{\infty}\operatorname{arccot}\frac{1-k^2+k^4}{2k}$

1 Answers1