Sum is always positive (prove): ∑_(k=0 to 2n) (−λ)^k (Γ(2n+1−k+kα))/(Γ(kα+1)Γ(2n+1−k))

CesarRocha

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Dec 11, 2023
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I tried to prove that the following sum is always positive, k=02n(λ)kΓ(2n+1k+kα)Γ(kα+1)Γ(2n+1k)\sum_{k=0}^{2n} (-\lambda)^k \frac{\Gamma{(2n+1-k+k\alpha)}}{\Gamma{(k\alpha+1)}\Gamma{(2n+1-k)}}
k=02n(λ)kΓ(2n+1k+kα)kαΓ(kα)Γ(2n+1k)=k=02n(λ)kkαB(kα,2n+1k)=k=02n(λ)kkα01tkα1(1t)2nkdt\sum_{k=0}^{2n} (-\lambda)^k \frac{\Gamma{(2n+1-k+k\alpha)}}{k\alpha\Gamma{(k\alpha)}\Gamma{(2n+1-k)}}=\sum_{k=0}^{2n} \frac{(-\lambda)^k}{k\alpha B(k\alpha,2n+1-k)}=\sum_{k=0}^{2n} \frac{(-\lambda)^k}{k\alpha \int_0^1 t^{k\alpha-1}(1-t)^{2n-k}dt}
 
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