The question is to express $$ \arctan a_1+ \arctan a_2 + \dots + \arctan a_n $$ as $\arctan(\cdot)$ I tried using the formula for the corresponding $\tan$ series but couldn't generalise to all cases.The principal branch is taken as $(-\pi/2,\pi/2)$.
Any ideas?
Use the addition formula $$\arctan a_1 + \arctan a_2=\arctan\left(\frac{a_1+a_2}{1-a_1a_2}\right)\quad (\text{mod}\,\pi)$$ In general, working in modulo $\pi$:
If $n=2k, k\in\mathbb{N}$: $$\sum_{n=1}^{2k}\arctan a_n=\arctan\left({\sum_{j=1}^k\left[(-1)^{j-1}\left(\sum_{\text{cyc}}\prod_{i=1}^{2j-1}a_i\right)\right]}\over{1-\sum_{j=1}^k\left[(-1)^{j-1}\left(\sum_{\text{cyc}}\prod_{i=1}^{2j}a_i\right)\right]}\right)$$
If $n=2k+1, k\in\mathbb{N}$: $$\sum_{n=1}^{2k+1}\arctan a_n=\arctan\left(\frac{\sum_{j=0}^k\left[(-1)^j\left(\sum_{\text{cyc}}\prod_{i=1}^{2j+1}a_i\right)\right]}{1-\sum_{j=1}^k\left[(-1)^j\left(\sum_{\text{cyc}}\prod_{i=1}^{2j}a_i\right)\right]}\right)$$
Hint: Let $z_n=1+ia_n$ then $\arctan a_n=\arg z_n$ and $$\arctan a_1+ \arctan a_2 + \dots + \arctan a_n=\arg(\prod_{k=1}^n z_k)=\arctan \dfrac{{\bf Im}\ {w}}{{\bf Re}\ {w}}$$ where $\displaystyle w=\prod_{k=1}^n z_k$.
Hint:
From the addition formula,
$$\tan(\arctan a+\arctan b+\arctan c)=\frac{\dfrac{a+b}{1-ab}+c}{1-\dfrac{a+b}{1-ab}c} =\dfrac{a+b+c-abc}{1-ab-bc-ca}$$
and
$$\tan(\arctan a+\arctan b+\arctan c+\arctan d) \\=\frac{{\dfrac{a+b+c-abc}{1-ab-bc-ca}}+d}{1-\dfrac{a+b+c-abc}{1-ab-bc-ca}d} \\=\dfrac{a+b+c+d-abc-bcd-cda-dab}{1-ab-bc-cd-da-ac-bd-ca+abcd}.$$
A regular pattern seems to emerge. We can write a recurrent form
$$\tan\left(\arctan\frac{p_{n-1}}{q_{n-1}}+\arctan a_n\right)=\frac{\dfrac{p_{n-1}}{q_{n-1}}+a_n}{1-\dfrac{p_{n-1}}{q_{n-1}}a_n}=\frac{p_{n-1}+q_{n-1}a_n}{q_{n-1}-p_{n-1}a_n}$$
and
$$\begin{cases}p_n=p_{n-1}+q_{n-1}a_n,\\q_n=q_{n-1}-p_{n-1}a_n.\end{cases}$$
From an operational point of view, this is equivalent to Nosrati's formula, where the product will be computed incrementally. The fully expanded formulas are complicated and computationally inefficient.