An infinite product inside a series

Calculus Level 5

For each positive integer m m we define the function P ( m ) : = ∏ k = 1 ∞ ( 1 − ( 2 m ) 2 ( 2 k − 1 ) 2 ) . P(m):=\prod_{k=1}^\infty\left(1-\dfrac{(2m)^2}{(2k-1)^2}\right). If ∑ m = 1 ∞ P ( m ) m 2 + 2 m − 1 m 2 − 2 m 3 = a b π + c d π 2 \sum_{m=1}^\infty P(m)\dfrac{m^2+2m-1}{m^2-2m^3}=\dfrac ab \pi+\dfrac cd \pi^2 for some positive integers a , b , c a,b,c and d d such that gcd ⁡ ( a , b ) = gcd ⁡ ( c , d ) = 1 \gcd(a,b)=\gcd(c,d)=1 , find a + b + c + d a+b+c+d .


The answer is 18.

Diego G by Diego G , 27, Spain

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1 solution

Define the infinite cosine product by : cos ⁡ ( π x ) = ∏ k = 1 ∞ ( 1 − 4 x 2 ( 2 k − 1 ) 2 ) \displaystyle \cos (\pi x) = \prod_{k=1}^{\infty} (1-\frac{4x^2}{(2k-1)^2}) and in that sense, P ( m ) = cos ⁡ ( π m ) = ( − 1 ) m P(m)=\cos(\pi m)=(-1)^m

Since, m 2 + 2 m − 1 m 2 − 2 m 3 = − 1 m 2 − 1 2 m − 1 \displaystyle \frac{m^2+2m-1}{m^2-2m^3} = -\frac{1}{m^2}-\frac{1}{2m-1}

Our sum S = ∑ m ≥ 1 P ( m ) m 2 + 2 m − 1 m 2 − 2 m 3 = ∑ m ≥ 1 ( − 1 ) m − 1 ( 1 m 2 + 1 2 m − 1 ) \displaystyle S=\sum_{m\ge 1}P(m)\frac{m^2+2m-1}{m^2-2m^3} = \sum_{m\ge 1}(-1)^{m-1} (\frac{1}{m^2}+\frac{1}{2m-1})

So, S = ∑ m = 1 ∞ ( − 1 ) m − 1 m 2 + ( − 1 ) m − 1 2 m − 1 = π 2 12 + π 4 \displaystyle S=\sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m^2}+\frac{(-1)^{m-1}}{2m-1} = \frac{\pi^2}{12}+\frac{\pi}{4} and thus making the answer 1 + 4 + 1 + 12 = 18 \boxed{1+4+1+12=18}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Well done! If one does not recall the product representation of the cosine function, a way is to define the sequence a n : = ∏ k = 1 n ( 1 − ( 2 m ) 2 ( 2 k − 1 ) 2 ) a_n:=\prod_{k=1}^n\left(1-\frac{(2m)^2}{(2k-1)^2}\right) and deduce that a n = ( − 1 ) m Γ ( 1 2 + m + n ) Γ ( 1 2 − m + n ) Γ ( 1 2 + n ) 2 . a_n=\frac{(-1)^m\Gamma(\tfrac12+m+n)\Gamma(\tfrac12-m+n)}{\Gamma(\tfrac12+n)^2}. Thus, using Stirling approximation it follows that a n → ( − 1 ) m a_n\to(-1)^m .

Diego G - 4 years, 9 months ago

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