Vtyjkyuk

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?

Hint: Let $z_n=1+ia_n$ then $arctan a_n=arg z_n$ and
$$arctan a_1+ arctan a_2 + dots + arctan a_n=arg(prod_k=1^n z_k)=arctan dfracbf Im wbf Re w$$
where $displaystyle w=prod_k=1^n z_k$.

Use the addition formula $$arctan a_1 + arctan a_2=arctanleft(fraca_1+a_21-a_1a_2right)quad (textmod,pi)$$ In general, working in modulo $pi$:

If $n=2k, kinmathbbN$: $$sum_n=1^2karctan a_n=arctanleft(sum_j=1^kleft[(-1)^j-1left(sum_textcycprod_i=1^2j-1a_iright)right]over1-sum_j=1^kleft[(-1)^j-1left(sum_textcycprod_i=1^2ja_iright)right]right)$$

If $n=2k+1, kinmathbbN$: $$sum_n=1^2k+1arctan a_n=arctanleft(fracsum_j=0^kleft[(-1)^jleft(sum_textcycprod_i=1^2j+1a_iright)right]1-sum_j=1^kleft[(-1)^jleft(sum_textcycprod_i=1^2ja_iright)right]right)$$

Hint:

From the addition formula,

$$tan(arctan a+arctan b+arctan c)=fracdfraca+b1-ab+c1-dfraca+b1-abc
=dfraca+b+c-abc1-ab-bc-ca$$

and

$$tan(arctan a+arctan b+arctan c+arctan d)
\=fracdfraca+b+c-abc1-ab-bc-ca+d1-dfraca+b+c-abc1-ab-bc-cad
\=dfraca+b+c+d-abc-bcd-cda-dab1-ab-bc-cd-da-ac-bd-ca+abcd.$$

A regular pattern seems to emerge. We can write a recurrent form

$$tanleft(arctanfracp_n-1q_n-1+arctan a_nright)=fracdfracp_n-1q_n-1+a_n1-dfracp_n-1q_n-1a_n=fracp_n-1+q_n-1a_nq_n-1-p_n-1a_n$$

and

$$begincasesp_n=p_n-1+q_n-1a_n,\q_n=q_n-1-p_n-1a_n.endcases$$

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.