CesarRocha
New member
- Joined
- Dec 11, 2023
- Messages
- 3
I tried to prove that the following sum is always positve,
[math]\sum_{k=0}^{2n} (-\lambda)^k \frac{\Gamma{(2n+1-k+k\alpha)}}{\Gamma{(k\alpha+1)}\Gamma{(2n+1-k)}}[/math]I concluded that [math]\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}[/math]but i don't know how to continue from here.
Thank in advance!
[math]\sum_{k=0}^{2n} (-\lambda)^k \frac{\Gamma{(2n+1-k+k\alpha)}}{\Gamma{(k\alpha+1)}\Gamma{(2n+1-k)}}[/math]I concluded that [math]\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}[/math]but i don't know how to continue from here.
Thank in advance!
