Factorials Fun

Calculus Level 3

1 3 ! + 2 5 ! + 3 7 ! + 4 9 ! + ⋯ = ? \large \dfrac { 1 }{ 3! } +\dfrac { 2 }{ 5! } +\dfrac { 3 }{ 7! } +\dfrac { 4 }{ 9! } +\cdots = \, ?

1 2 \frac 12 1 2 e \frac 1{2e} 1 3 e \frac 1{3e} 1 1 1 e \frac 1e

Achal Jain by Achal Jain , 19, India

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1 solution

Chew-Seong Cheong
Sep 24, 2016

Consider the following,

e − 1 = 1 − 1 1 ! + 1 2 ! − 1 3 ! + 1 4 ! − 1 5 ! + 1 6 ! − 1 7 ! + 1 8 ! − 1 9 ! + . . . 1 e = 0 + 2 3 ! + 4 5 ! + 6 7 ! + 8 9 ! + . . . ⟹ 1 2 e = 1 3 ! + 2 5 ! + 3 7 ! + 4 9 ! + . . . \begin{aligned} e^{-1} & = \color{#3D99F6}{1 - \frac 1{1!}} + \color{#D61F06}{\frac 1{2!} - \frac 1{3!}} + \color{#3D99F6}{\frac 1{4!} - \frac 1{5!}} + \color{#D61F06}{\frac 1{6!} - \frac 1{7!}} + \color{#3D99F6}{\frac 1{8!} - \frac 1{9!}} +... \\ \frac 1e & = \color{#3D99F6}{0} + \color{#D61F06}{\frac 2{3!}} + \color{#3D99F6}{\frac 4{5!}} + \color{#D61F06}{\frac 6{7!}} + \color{#3D99F6}{\frac 8{9!}} +... \\ \implies \boxed{\dfrac 1{2e}} & = \frac 1{3!} + \frac 2{5!} + \frac 3{7!} + \frac 4{9!} +... \end{aligned}

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Is this from the Taylor series expansion of e^x, where -1 is substituted?

John Frank - 4 years, 8 months ago

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Yes. e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + . . . e^x = 1 + x + \dfrac {x^2}{2!} + \dfrac {x^3}{3!} + \dfrac {x^4}{4!} + ... . Substituting x = − 1 x=-1 , we get: e − 1 = 1 − 1 + 1 2 ! − 1 3 ! + 1 4 ! + . . . e^{-1} = 1 -1 + \dfrac 1{2!} - \dfrac 1{3!} + \dfrac 1{4!} + ... .

Chew-Seong Cheong - 4 years, 8 months ago

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