Alexander Alenitsyn
Высший разум
(760028)
17 лет назад
It is more convenient to write the index using the symbol _ (for instance, b_n, instead of b\n)
1) Correction of the hint: the sum is equal to b_1-lim b_n, not b_1-lim A_n
2) - arctan(1/(2n^2)) = arctan p_n - arctan p_(n+1), where p_n = 2n -1. You can check it taking tan of the left and right sides of this equality and applying the formula tan(x - y) = (tan x - tan y)/(1 + tan x*tan y).
3) Denote b_n = arctan p_n and get the desired! The answer: b1 - arctan 1 = pi/2 - pi/4 = pi/4.
In case you wish more details, send me e-mail.
Yours sincerely, A.Alenitsyn
A\n=arctan(1/2n^2).
(Здесь указание. Нужно представить общий член в виде b/n-b/n+1. Тогда сумма ряда=b1-B, где В=lim A\n.)