Sum and Product ...

Calculus Level pending

∑ m = 1 ∞ ( lim k → m ∏ n = 1 ∞ n 2 − k 2 e 2 k − 1 n ( − 2 k + n + 1 ) n ) = 2 e γ ( b cos ⁡ ( e − γ ) − a ) + c sin ⁡ ( e − γ ) \sum _{m=1}^{\infty } \left(\underset{k\to m}{\text{lim}}\prod _{n=1}^{\infty } \frac{n^2-k^2}{e^{\frac{2 k-1}{n}} (-2 k+n+1) n}\right)=2 e^{\gamma } \left(b \cos \left(e^{-\gamma }\right)-a\right)+c \sin \left(e^{-\gamma }\right)

for positive integers a , b , c a,b,c . Submit a + b + c a+b+c .


The answer is 4.

Christopher Criscitiello by Christopher Criscitiello , 25, USA

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1 solution

HInt: Messing around with the functional equation for the zeta function, I found that

∏ n = 1 ∞ 2 ( 2 ( 1 − n ) n ( 2 n − s ) ( 2 n + s ) − s 2 n + s + 1 ) e s − 1 n = π 2 s π s − 1 s e − γ ( 1 − s ) ζ ( 1 − s ) 4 ( 1 − s ) ζ ( s ) \prod _{n=1}^{\infty } 2 \left(\frac{2 (1-n) n}{(2 n-s) (2 n+s)}-\frac{s}{2 n+s}+1\right) e^{\frac{s-1}{n}}=\frac{\pi 2^s \pi ^{s-1} s e^{-\gamma (1-s)} \zeta (1-s)}{4 (1-s) \zeta (s)}

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