Fun with 2018 #4

Calculus Level 5

∑ n = 1 ∞ ( − 1 ) n − 1 2 n ( 2 n − 1 ) ( 1 2018 ) 2 n = 1 2018 arctan ⁡ ( 1 2018 ) − 1 2 log ⁡ ( 1 + 1 A ) \large \sum_{n = 1}^\infty \frac {(-1)^{n - 1}}{2n(2n - 1)}\left (\frac{1}{2018} \right )^{2n} = \frac{1}{2018} \arctan \left (\frac{1}{2018} \right ) - \frac{1}{2}\log \left(1 + \frac{1}{A}\right)

What is the value of A A such that the equation above is satisfied?


The answer is 4072324.

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Guillermo Templado by Guillermo Templado , 46, Spain

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2 solutions

Relevant wiki: Maclaurin Series

S = ∑ n = 1 ∞ ( − 1 ) n − 1 x 2 n 2 n ( 2 n − 1 ) where x = 1 2018 = ∑ n = 1 ∞ ( − 1 ) n − 1 x 2 n ( 1 2 n − 1 − 1 2 n ) = x ∑ n = 1 ∞ ( − 1 ) n − 1 x 2 n − 1 2 n − 1 − 1 2 ∑ n = 1 ∞ ( − 1 ) n − 1 x 2 n n By Maclaurin series = x arctan ⁡ x − 1 2 log ⁡ ( 1 + x 2 ) Putting back x = 1 2018 = 1 2018 arctan ⁡ ( 1 2018 ) − 1 2 log ⁡ ( 1 + 1 201 8 2 ) \begin{aligned} S & = \sum_{n=1}^\infty \frac {(-1)^{n-1}{\color{#3D99F6}x}^{2n}}{2n(2n-1)} & \small \color{#3D99F6} \text{where }x = \frac 1{2018} \\ & = \sum_{n=1}^\infty (-1)^{n-1}x^{2n}\left(\frac 1{2n-1} - \frac 1{2n}\right) \\ & = x \sum_{n=1}^\infty \frac {(-1)^{n-1}x^{2n-1}}{2n-1} - \frac 12 \sum_{n=1}^\infty \frac {(-1)^{n-1}x^{2n}}n & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = x \arctan x - \frac 12 \log \left(1+x^2\right) & \small \color{#3D99F6} \text{Putting back }x = \frac 1{2018} \\ & = \frac 1{2018} \arctan \left(\frac 1{2018}\right) - \frac 12 \log \left(1+\frac 1{2018^2} \right) \end{aligned}

Therefore, A = 201 8 2 = 4072324 A = 2018^2 = \boxed{4072324} .

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∑ n = 1 ∞ ( − 1 ) n − 1 2 n − 1 x 2 n − 1 = arctan (x) , ∀ x ∈ R \displaystyle \sum_{n = 1}^\infty \frac{(-1)^{n -1}}{2n - 1} x^{2n - 1} = \text {arctan (x)}, \forall \space x \in \mathbb{R} such that ∣ x ∣ ≤ 1 |x| \leq 1 . Hence, integrating ∑ n = 1 ∞ ( − 1 ) n − 1 2 n ( 2 n − 1 ) x 2 n = ∫ 0 x arctan (t) d t = x ⋅ arctan (x) − 1 2 log ⁡ ( 1 + x 2 ) , ∀ x ∈ R \displaystyle \sum_{n = 1}^\infty \frac{(-1)^{n -1}}{2n(2n - 1)} x^{2n} = \int_{0}^x \text {arctan (t)} \, dt = x \cdot \text {arctan (x)} - \frac{1}{2} \log (1 + x^2), \forall \space x \in \mathbb{R} as long as ∣ x ∣ ≤ 1 |x| \leq 1 . Substituing x = 1 2018 x = \frac{1}{2018} we get A = 201 8 2 = 4072324 A = 2018^2 = 4072324

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