1-prove-that-arctan-a-arctanb-arctan-a-b-1-ab-with-ab-1-2-find-the-value-of-S-N-n-1-N-1-n-arctan-2n-1-n-2-n-1-

Question Number 57848 by Abdo msup. last updated on 13/Apr/19

1) p r o v e t h a t a r c t a n (a) + a r c t a n b = a r c t a n (\frac{a + b}{1 - a b})

w i t h a b \neq 1

2) f i n d t h e v a l u e o f S_{N} = \sum_{n = 1}^{N} {(- 1)}^{n} a r c t a n (\frac{2 n + 1}{n^{2} + n - 1})

Commented by maxmathsup by imad last updated on 13/Apr/19

1) l e t p u t α = a r c t a n (a) a n d β = a r c t a n (b) \Rightarrow

t a n (α + β) = \frac{t a n (α) + t a n (β)}{1 - t a n (α) t a n (β)} = \frac{a + b}{1 - a b} (w e s u p p o s e a b \neq 1) \Rightarrow

α + β = a r c t a n (\frac{a + b}{1 - a b})

2) w e h a v e S_{n} = - \sum_{n = 1}^{N} {(- 1)}^{n} a r c t a n (\frac{2 n + 1}{1 - n (n + 1)})

= - \sum_{n = 1}^{N} {(- 1)}^{n} a r c t a n (\frac{n + n + 1}{1 - n (n + 1)}) = - \sum_{n = 1}^{N} {(- 1)}^{n} {a r c t a n (n) + a r c t a n (n + 1)}

\Rightarrow - S_{n} = - a r c t a n (1) - a r c t a n (2) + a r c t a n (2) + a r c t a n (3) - \dots

+ {(- 1)}^{n} a r c t a n (n) + {(- 1)}^{n} a r c t a n (n + 1)

= {(- 1)}^{n} a r c t a n (n + 1) - \frac{π}{4} \Rightarrow S_{N} = \frac{π}{4} - {(- 1)}^{n} a r c t a n (n + 1) .

Commented by maxmathsup by imad last updated on 13/Apr/19

S_{N} = \frac{π}{4} - {(- 1)}^{N} a r c t a n (N + 1) .