Algebra under the Calculus

Calculus Level 3

Simplify ln ⁡ ( 1 2 ! − 1 3 ! + 1 4 ! − 1 5 ! + ⋯ ) \ln \left(\frac 1{2!} - \frac 1{3!} + \frac 1{4!} - \frac 1{5!} + \cdots \right)


The answer is -1.

Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

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3 solutions

Paul Ryan Longhas
Feb 19, 2015

By Maclaurin Series for e x e^x e − 1 = ( 1 / 2 ! ) − ( 1 / 3 ! ) + ( 1 / 4 ! ) . . . e^{-1} = (1/2!)-(1/3!)+(1/4!)... Hence, l n e − 1 ln e^{-1} = > − 1 => -1

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Should this be a level 4 problem? I think not.

Prasun Biswas - 6 years, 3 months ago
Chew-Seong Cheong
Sep 13, 2018

Note that:

e x = 1 + x 1 ! + x 2 2 ! + x 3 3 ! + x 4 4 ! + x 5 5 ! + ⋯ Putting x = − 1 ⟹ e − 1 = 1 − 1 + 1 2 ! − 1 3 ! + 1 4 ! − x 5 5 ! + ⋯ \begin{aligned} e^x & = 1 + \frac x{1!} + \frac {x^2}{2!} + \frac {x^3}{3!} + \frac {x^4}{4!} + \frac {x^5}{5!} + \cdots & \small \color{#3D99F6} \text{Putting }x=-1 \\ \implies e^{-1} & = 1 - 1 + \frac 1{2!} - \frac 1{3!} + \frac 1{4!} - \frac {x^5}{5!} + \cdots \end{aligned}

Therefore, ln ⁡ ( 1 2 ! − 1 3 ! + 1 4 ! − x 5 5 ! + ⋯ ) = ln ⁡ ( e − 1 ) = − 1 \displaystyle \ln \left(\frac 1{2!} - \frac 1{3!} + \frac 1{4!} - \frac {x^5}{5!} + \cdots\right) = \ln \left(e^{-1}\right) = \boxed{-1} .

2 Helpful 1 Interesting 1 Brilliant 0 Confused
Fox To-ong
Mar 24, 2015

using maclaurin series for e^x

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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