I would like to find the upper bound (or simplification) of this expression: $$\sum_{j=1}^{n+1}\sum_{i=0}^{j-1} a^{j+i} {j+i \choose i}{n+1\choose j}{n \choose i}/{2n+1 \choose j+i}$$ where $0<a<1$ is constant.
Please help me to solve it. Thanks a lot.
We can write our sum in the more symmetric fashion: $$\sum_{j=1}^{n+1}\sum_{i<j}a^{i+j}\frac{\binom{n+1}{i}\binom{n+1}{j}}{\binom{2n+2}{i+j}}\cdot\left(1+\frac{j-i}{2n+2-(i+j)}\right)$$ and see that it is bounded by the symmetric one:
$$\sum_{j=1}^{n+1}\sum_{i<j}a^{i+j}\frac{\binom{n+1}{i}\binom{n+1}{j}}{\binom{2n+2}{i+j}}\cdot\left(1+\frac{n+1}{2n+2-(i+j)}\right)$$ that can be bounded by: $$\sum_{k=1}^{2n+1}\frac{a^k}{\binom{2n+2}{k}}\left(1+\frac{n+1}{2n+2-k}\right)\frac{1}{2}\binom{2n+2}{k}$$ that can be bounded by: $$\frac{1}{2(1-a)}+\frac{(n+1)a^{2n+2}}{2}\sum_{k=1}^{2n+1}\frac{a^{-k}}{k}$$ that can be bounded by: $$\frac{1}{2(1-a)}+\frac{n+1}{2}a\, H_{2n+1}.$$