Crazy Sum.

Calculus Level 5

n = 1 r = 0 n 1 ( π r ) 2 n n ! = ( A B ) π C \large \sum_{n=1}^{\infty}\frac{\prod_{r=0}^{n-1} (\pi-r)}{2^{n}n!} = \left(\frac{A}{B}\right)^{\pi}-C

If the equation above holds true for positive coprime integers A A , B B , and C C , enter the value of A + B C A+B-C .


The answer is 4.

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2 solutions

敬全 钟
Mar 28, 2018

This problem can be solved easily using binomial series, as follows. ( 1 + x ) π = n = 0 ( π n ) x n = 1 + n = 1 r = 0 n 1 ( π r ) n ! x n . \begin{aligned} (1+x)^{\pi}&=&\sum^{\infty}_{n=0}\binom{\pi}{n}x^n\\ &=&1+\sum^{\infty}_{n=1}\frac{\prod^{n-1}_{r=0}(\pi-r)}{n!}x^n. \end{aligned} The answer follows by substituting x = 1 2 . x=\frac{1}{2}.

Can we use combination over irrational numbers like π \pi as n n should be an integer?

Sahil Silare - 3 years, 2 months ago

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Indeed, we can. This is known as the generalized binomial coefficient, which has the definition ( α k ) = α ( α 1 ) ( α 2 ) . . . ( α k + 1 ) k ! \binom{\alpha}{k}=\frac{\alpha(\alpha-1)(\alpha-2)...(\alpha-k+1)}{k!} for any arbitrary number (negative, real or even complex) α \alpha and positive integer k . k.

敬全 钟 - 3 years, 2 months ago
Vijay Simha
Mar 27, 2018

Is the answer (3/2)^pi - 1?

I got this by using Gamma functions

Write full solutions please.

Sahil Silare - 3 years, 2 months ago

I also got by gamma function

A Former Brilliant Member - 3 years, 2 months ago

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