Question: \begin{align} \sum_{k=1}^{n} \arctan(k)=? \end{align}
My approach:
\begin{align} \prod_{k=1}^{n} (1+ik)=a+ib\qquad(1)\end{align} \begin{align} \Rightarrow arg \Bigg(\prod_{k=1}^{n} (1+ik)\Bigg) =arg \Big(a+bi\Big) \end{align} \begin{align} \Rightarrow \sum_{k=1}^{n} \arctan(k) =arg \Big(a+bi\Big) \end{align}
I am stuck here because $ (1) $ is too big to solve by hand and get $ a,b $ . Can anyone help find it out? Also I have a kinda similar problem that I have faced when tried it to solve by hand and get $ a,b $. Here's the link Sum of $ ^nC_k $ terms from $ n $ numbers and each term having $ k $ numbers multiplied .