Mistaken Identity

Calculus Level 3

n = ( ∑ r = 0 ∞ 1 r ! ) 8 i ∑ k = 1 ∞ ( − 1 ) k + 1 2 k − 1 \large n = \left(\sum_{r=0}^\infty \dfrac1{r!} \right)^{8i \sum_{k=1}^\infty \frac{ (-1)^{k+1}}{2k-1}}

Evaluate n n to 3 significant figures.

Notation : i = − 1 i=\sqrt{-1} is the imaginary unit .


The answer is 1.0.

William Whitehouse by William Whitehouse , 22, United Kingdom

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2 solutions

n = ( ∑ r = 0 ∞ 1 r ! ) 8 i ∑ k = 1 ∞ ( − 1 ) k + 1 2 k − 1 Using Maclaurin series = e 8 i tan ⁡ − 1 ( 1 ) = e 8 i ⋅ π 4 = e 2 π i By Euler’s formula = cos ⁡ 2 π + i sin ⁡ 2 π = 1 \large \begin{aligned} n & = \left(\color{#3D99F6}{\sum_{r=0}^\infty \dfrac1{r!}} \right)^{8i \color{#3D99F6}{\sum_{k=1}^\infty \frac{ (-1)^{k+1}}{2k-1}}} & \small \color{#3D99F6}{\text{Using Maclaurin series}} \\ & = \color{#3D99F6}{e}^{8i\color{#3D99F6}{\tan^{-1}(1)}} \\ & = e^{8i \cdot \frac \pi 4} \\ & = \color{#3D99F6}{e^{2\pi i}} & \small \color{#3D99F6}{\text{By Euler's formula}} \\ & = \color{#3D99F6}{\cos 2\pi + i \sin 2 \pi} \\ & = \boxed{1} \end{aligned}

References:

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The first part of the problem (the sum of reciprocal factorials) is a formula for e, the mathematical constant. The exponent is the Gregory series which evaluates to π 4 \frac{\pi}{4} . This means that the whole expression evaluates to e 2 i π e^{2i\pi} which is not Euler's identity (hence the name of the problem) but that squared, ( − 1 ) 2 = 1 (-1)^{2}=1

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