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Calculus Level 3

( 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! + . . . ) arcsin ⁡ ( − 1 ) = ? \large \left(\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ...\right)^{\arcsin(-1)} = \ ?

e − i e^{-i} − i -i i i i^{i} e e

Hobart Pao by Hobart Pao , 23, USA

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1 solution

We know that ∑ k = 0 ∞ 1 k ! = e \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{k!} = e and that arcsin ⁡ ( − 1 ) = − π 2 . \arcsin(-1) = -\dfrac{\pi}{2}.

Thus the given expression is equal to e − π 2 . \large e^{-\frac{\pi}{2}}.

Now i = e i π 2 , \large i = e^{\frac{i\pi}{2}}, so i i = e i 2 π 2 = e − π 2 . \large i^{i} = e^{\frac{i^{2}\pi}{2}} = e^{-\frac{\pi}{2}}.

Thus the given expression is in fact equal to i i . \large \boxed{i^{i}}.

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