Multiplication

Calculus Level 3

The value of the product ( 1 + 1 1 ! + 1 2 ! + ⋯ ) ( 1 − 1 1 ! + 1 2 ! − 1 3 ! + ⋯ ) \left(1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \cdots\right) \left(1 - \dfrac{1}{1!} + \dfrac{1}{2!} -\dfrac{1}{3!} + \cdots\right) is

0 e 2 e^2 1 ln ⁡ 2 \ln 2

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Ravneet Singh by Ravneet Singh , 30, India

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1 solution

In the taylor expansion of e x = 1 + x + x 2 2 ! + ⋯ \displaystyle e^x=1+x+\dfrac{x^2}{2!}+\cdots , put x = 1 , − 1 x=1,-1 successively and hence obtain that the product is equal to ( e ) × ( e − 1 ) = 1 \displaystyle (e)\times(e^{-1})=\boxed{1}

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