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Calculus Level 2

4 ∑ n = 1 ∞ ( ( − 1 ) n + 1 2 n − 1 ) = ? \large{4\sum_{n=1}^{\infty} \bigg ( \frac{(-1)^{n+1}}{2n-1} \bigg ) = \ ?}

This is not an original problem. \small{\text{This is not an original problem.}}

e π \pi 5 \sqrt{5} ϕ \phi None of These

Alex Delhumeau by Alex Delhumeau , 20, USA

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1 solution

Eamon Gupta
Aug 14, 2015

4 ∑ n = 1 ∞ ( ( − 1 ) n + 1 2 n − 1 ) = 4 ( 1 − 1 3 + 1 5 − 1 7 ⋯ ) \large{4\sum_{n=1}^{\infty}(\frac{(-1)^{n+1}}{2n-1}) = 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\cdots)}

Gregory's series is defined as

arctan ⁡ ( x ) = x − x 3 3 + x 5 5 − x 7 7 ⋯ \arctan(x) = \large{ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} \cdots}

By substituting x = 1:

1 − 1 3 + 1 5 − 1 7 ⋯ = arctan ⁡ ( 1 ) \large{ 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} \cdots} = \arctan(1)

4 arctan ⁡ ( 1 ) = π \large{4\arctan(1) = \boxed{\pi}}

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