What is the summation of the finite sequence: $$\sum\limits_{i = 1}^n {\frac{1}{i}\left( {\begin{array}{*{20}{c}} {2i - 2}\\ {i - 1} \end{array}} \right)\left( {\begin{array}{*{20}{c}} {2n + 2 - 2i}\\ {n + 1 - i} \end{array}} \right)} $$
My professor leave us the answer: $\frac{{2n}}{{n + 1}}\left( {\begin{array}{*{20}{c}} {2n}\\ n \end{array}} \right)$, if I remembered it right.
But why such a complex summation has such a neat closed form expression? I'm stupid:(
Hint. You may use the same technique exposed here.
From $$ \sum_{i=0}^{\infty} \binom{2i}i x^i=\frac{1}{\sqrt{1-4x}},\quad |x|<\frac14, \tag1 $$ by integration, you get $$ \sum_{i=1}^{\infty} \frac1i\binom{2i-2}{i-1} x^i=\frac{1}{2}\left(1-\sqrt{1-4x} \right),\quad |x|<\frac14. \tag2 $$ and then use the Cauchy product.