Double Trouble!

Calculus Level 3

Find the nearest integer to ∑ n = 0 ∞ 1 n ! ! \displaystyle \sum_{n=0}^{\infty}{\frac{1}{n!!}} , where n ! ! n!! is the double factorial .


The answer is 3.

Kartik Sharma by Kartik Sharma , 21, India

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2 solutions

Chew-Seong Cheong
Aug 30, 2018

Note that ∑ n = 0 ∞ 1 n ! ! = ∑ n = 0 ∞ 1 ( 2 n ) ! ! + ∑ n = 0 ∞ 1 ( 2 n + 1 ) ! ! \displaystyle \sum_{n=0}^\infty \frac 1{n!!} = \sum_{n=0}^\infty \frac 1{(2n)!!} + \sum_{n=0}^\infty \frac 1{(2n+1)!!} and also that ∑ n = 0 ∞ 1 ( 2 n + 1 ) ! ! < ∑ n = 0 ∞ 1 ( 2 n ) ! ! \displaystyle \sum_{n=0}^\infty \frac 1{(2n+1)!!} < \sum_{n=0}^\infty \frac 1{(2n)!!} and ∑ n = 0 ∞ 1 ( 2 n + 1 ) ! ! > ∑ n = 0 ∞ 1 ( 2 n + 1 ) ! \displaystyle \sum_{n=0}^\infty \frac 1{(2n+1)!!} > \sum_{n=0}^\infty \frac 1{(2n+1)!} . Therefore,

∑ n = 0 ∞ 1 ( 2 n ) ! ! + ∑ n = 0 ∞ 1 ( 2 n + 1 ) ! < ∑ n = 0 ∞ 1 n ! ! < 2 ∑ n = 0 ∞ 1 ( 2 n ) ! ! ∑ n = 0 ∞ 1 2 n n ! + ∑ n = 0 ∞ 1 ( 2 n + 1 ) ! < ∑ n = 0 ∞ 1 n ! ! < 2 ∑ n = 0 ∞ 1 2 n n ! e + sinh ⁡ 1 < ∑ n = 0 ∞ 1 n ! ! < 2 e 2.823922464 < ∑ n = 0 ∞ 1 n ! ! < 3.297442541 ⟹ ∑ n = 0 ∞ 1 n ! ! ≈ 3 \begin{aligned} \sum_{n=0}^\infty \frac 1{(2n)!!} + \sum_{n=0}^\infty \frac 1{(2n+1)!} < & \sum_{n=0}^\infty \frac 1{n!!} < 2 \sum_{n=0}^\infty \frac 1{(2n)!!} \\ \sum_{n=0}^\infty \frac 1{2^nn!} + \sum_{n=0}^\infty \frac 1{(2n+1)!} < & \sum_{n=0}^\infty \frac 1{n!!} < 2 \sum_{n=0}^\infty \frac 1{2^nn!} \\ \sqrt e + \sinh 1 < & \sum_{n=0}^\infty \frac 1{n!!} < 2 \sqrt e \\ 2.823922464 < & \sum_{n=0}^\infty \frac 1{n!!} < 3.297442541 \\ \implies & \sum_{n=0}^\infty \frac 1{n!!} \approx \boxed 3 \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Aareyan Manzoor
Nov 10, 2015

we could use very hard calculus and what not to solve this or : S = ∑ n = 0 ∞ ( 1 n ! ! ) = 1 + 1 + 1 2 + 1 6 ! + 1 24 ! . . . . . S=\sum_{n=0}^\infty (\dfrac{1}{n!!})=1+1+\dfrac{1}{2}+\dfrac{1}{6!}+\dfrac{1}{24!}..... we easily deduce that 1 + 1 + 1 2 < S < ∑ n = 0 ∞ ( 1 n ! ) 1+1+\dfrac{1}{2}<S<\sum_{n=0}^\infty (\dfrac{1}{n!}) 2.5 < S < e ≈ 2.718 2.5<S<e\approx 2.718 ⌈ S ⌋ = 3 \lceil S\rfloor=3

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Damn! I framed it wrong. I wanted exact solution, the closed form, I mean which is e ( 1 + π 2 erf ( 1 2 ) ) \sqrt{e}\left(1 + \sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{1}{\sqrt{2}}\right)\right) . Try Triple Cripple too.

Kartik Sharma - 5 years, 7 months ago

You've actually misunderstood what a double factorial is, n ! ! n!! is not the same as ( n ! ) ! (n!)! . But don't let that take away from the beauty of your solution, it's a nice solution but not to this question.

Isaac Buckley - 5 years, 6 months ago

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