Factorials... This Reminds Me of Something..

Calculus Level 5

∑ n = 0 ∞ 1 ( n + 1 ) ! ( 1 2 − n ) ! \large \sum_{n=0}^{\infty} \frac{1}{(n+1)! \left(\frac{1}{2} - n\right)!}

The infinite sum above can be expressed as k a a − b c π m \displaystyle k\frac{a\sqrt{a}-b}{c}\pi^m , where a a , b b , c c , and k k are positive integers, k k and c c are coprime. Find k + a + b + c + m − 4. \displaystyle k+a+b+c+m -4.

Note: Factorials of real numbers that are not non-negative integers, is well defined using the Gamma function. We have x ! = Γ ( x + 1 ) x! = \Gamma(x+1) .

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The answer is 5.5.

Hasan Kassim by Hasan Kassim , 22, Lebanon

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1 solution

Hasan Kassim
Sep 19, 2014

First , look at the term ( 1 2 − n ) ! \displaystyle (\frac{1}{2} - n)! , It reminds us of the following expansion:

1 + x = ∑ m = 0 ∞ ( 1 2 m ) x m \displaystyle \sqrt{1+x} = \sum_{m=0}^{\infty} {\frac{1}{2} \choose m} x^m

So let's try to build up this in our sum:

I = ∑ n = 0 ∞ 1 ( n + 1 ) ! ( 1 2 − n ) ! = 1 ( 1 2 ) ! ∑ n = 0 ∞ ( 1 2 ) ! ( n ) ! ( 1 2 − n ) ! ( n + 1 ) \displaystyle I=\sum_{n=0}^{\infty} \frac{1}{(n+1)!(\frac{1}{2}-n)!} = \frac{1}{(\frac{1}{2})!} \sum_{n=0}^{\infty} \frac{(\frac{1}{2})!}{(n)!(\frac{1}{2}-n)!(n+1)}

= 1 ( 1 2 ) ! ∑ n = 0 ∞ ( 1 2 n ) 1 n + 1 \displaystyle = \frac{1}{(\frac{1}{2})!} \sum_{n=0}^{\infty} {\frac{1}{2} \choose n } \frac{1}{n+1}

Now define the function f ( x ) = 1 ( 1 2 ) ! ∑ n = 0 ∞ ( 1 2 n ) x n + 1 n + 1 \displaystyle f(x)= \frac{1}{(\frac{1}{2})!} \sum_{n=0}^{\infty} {\frac{1}{2} \choose n } \frac{x^{n+1}}{n+1}

= > f ′ ( x ) = 1 ( 1 2 ) ! ∑ n = 0 ∞ ( 1 2 n ) x n = 1 ( 1 2 ) ! 1 + x \displaystyle => f'(x)=\frac{1}{(\frac{1}{2})!} \sum_{n=0}^{\infty} {\frac{1}{2} \choose n } x^n =\frac{1}{(\frac{1}{2})!} \sqrt{1+x}

We know that I = f ( 1 ) − f ( 0 ) \displaystyle I= f(1)-f(0)

= ∫ 0 1 f ′ ( x ) d x = 1 ( 1 2 ) ! ∫ 0 1 1 + x d x = 1 ( 1 2 ) ! 2 3 ( x + 1 ) 3 2 ∣ 0 1 \displaystyle = \int_0^1 f'(x) dx =\frac{1}{(\frac{1}{2})!} \int_0^1 \sqrt{1+x} dx = \frac{1}{(\frac{1}{2})!} \frac{2}{3} \left. (x+1)^{\frac{3}{2}} \right |_0^1

Substituting and rearranging, we get:

I = 4 2 2 − 1 3 π − 1 2 \displaystyle \boxed{ I = 4\frac{2\sqrt{2} - 1}{3}\pi^{-\frac{1}{2}} }

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Not the way I did it. Long manipulations of the gamma function was my way but that is way too long to post. Nice work this is cleaner than mine!

A Former Brilliant Member - 6 years, 8 months ago

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Well if you can post your solution or just the main idea of it , I would be thankful :)

Hasan Kassim - 6 years, 8 months ago

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