telescopic series

Question

$$\sum_{n=1}^\infty \left(\arctan(n+2)-\arctan(n)\right)$$

Wondering what the answer is to this. I Left over terms that I have are $-\arctan(1)$, $-\arctan(2)$, $\arctan(n+2)$ and $\arctan(n+1)$. Not sure if this is correct. Thanks

Answer 1

$$\sum_{n=1}^N (\arctan(n+2) -\arctan(n)) = \arctan(N+2) +\arctan(N+1) - \arctan(2) - \arctan(1) $$

Since $\arctan(x) \to \pi/2$ as $x \to \infty$, the infinite sum is $$ \pi - \arctan(2) - \arctan(1) $$

Answer 2

You're almost there. Note that

$$\sum_{k=1}^n\left( \arctan(k+2)-\arctan(k)\right)=\arctan(n+2)+\arctan(n+1)-\arctan(2)-\arctan(1)$$

Since $\lim_{n\to \infty}\arctan(n)=\pi/2$, then

$$\sum_{k=1}^\infty\left( \arctan(k+2)-\arctan(k)\right)=\pi/2+\pi/2-\arctan(2)-\pi/4=\frac{3\pi}{4}-\arctan(2)$$