Use the identity arctan 1/2 + arctan 1/3 = pi/4 to express pi as the sum of two infinite series.?

Question 1. Use the identity tan^(-1) 1/2 + tan^(-1) 1/3 = pi/4 to express pi as the sum of two infinite series.





Question 2. Use (11.48)(e) with x = 1 to represent pi as the sum of an infinite series.





(11.48)(e) states that tan^(-1) x = x - x^3/3 + x^5/5 - x^7/7 + ... + (-1)^n x^(2n+1) / (2n+1).





I think I should apply Taylor's series with x = 1/2 and 1/3 in two separate series. So I should take the derivative of arctan x, which is (if I remember correct) 1/(1+x^2), and then take the second derivative, and so on. If arctan x = f(x), then arctan x = f(x) + f'(x) + f''(x)/2! + f'''(x)/3! + ... I should check the remainder R_n (x) in Taylor's formula with remainder to ensure that f^(n+1) (z)/(n+1)! * (x-c)^(n+1) has a limit 0 as n approaches infinity.





I don't think the nth derivative of arctan x has any pattern (not a constant).

Use the identity arctan 1/2 + arctan 1/3 = pi/4 to express pi as the sum of two infinite series.?
There is a pattern in the Taylor expansion of arctan(x).


I will leave it to you to verify this, but here is the series:





arctan(x) = x - (1/3)x^3 + (1/5)x^5 - (1/7)x^7 ...





The series converges for -1 %26lt; x %26lt; 1

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