Summing up factorials (2)

Calculus Level 5

a = ∑ r = 1 ∞ ( − 1 ) r + 1 ( 2 r − 1 ) ! ! ( 2 r ) ! ! \large a= \sum_{ r=1 }^{ \infty }{ { (-1) }^{ r+1 } \dfrac{ ( 2r-1 )!! }{ ( 2r )!! } }

Find the closed form of a a . Submit your answer as ⌊ 100 a ⌋ \lfloor 100a \rfloor .

Notations:

  • ! ! !! denotes the double factorial notation. For example, 10 ! ! = 10 × 8 × 6 × 4 × 2 10!!=10\times8\times6\times4\times2 .

  • ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 29.

Rishi Sharma by Rishi Sharma , 32, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mark Hennings
Feb 1, 2017

Since the Taylor series expansion of ( 1 + x ) − 1 2 (1+x)^{-\frac12} is ( 1 + x ) − 1 2 = 1 + ∑ r = 1 ∞ ( − 1 ) r ( 2 r − 1 ) ! ! ( 2 r ) ! ! x r ∣ x ∣ < 1 (1 + x)^{-\frac12} \; = \; 1 + \sum_{r=1}^\infty (-1)^r \frac{(2r-1)!!}{(2r)!!} x^r \hspace{2cm} |x| < 1 the fact that the series a a converges allows us to calculate a = − ∑ r = 1 ∞ ( − 1 ) r ( 2 r − 1 ) ! ! ( 2 r ) ! ! = 1 − lim ⁡ x → 1 ( 1 + x ) − 1 2 = 1 − 1 2 a \; = \; -\sum_{r=1}^\infty (-1)^r \frac{(2r-1)!!}{(2r)!!} \; = \; 1 - \lim_{x \to 1} (1 + x)^{-\frac12} \; = \; 1 - \tfrac{1}{\sqrt{2}} and hence the answer is ⌊ 100 a ⌋ = ⌊ 50 ( 2 − 2 ) ⌋ = 29 \lfloor 100a \rfloor = \lfloor 50(2-\sqrt{2}) \rfloor = \boxed{29} .

3 Helpful 0 Interesting 0 Brilliant 1 Confused

Nice observation ..of taylor . i did same but man I forget (1) to add at -√0.5 thus leading to -71

Dhruv Joshi - 4 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...