<s>[INST] Prove $\sum_{i=0}^{n-1} \binom{2i}{i}\frac{1}{i+1}\frac{1}{2^{2i}} = 2(1 - \frac{1}{2^{2n}}\binom{2n}{n})$ How can I prove $\sum_{i=0}^{n-1} \binom{2i}{i}\frac{1}{i+1}\frac{1}{2^{2i}} = 2(1 - \frac{1}{2^{2n}}\binom{2n}{n})$ without induction? I think I have to use Taylor series at some point.
 [/INST] We may notice that by setting $C_i=\frac{1}{i+1}\binom{2i}{i}$ and $A_i=\frac{C_i}{4^i}$ we have
$$ f(x) = \sum_{i\geq 0}A_i x^i = 2\frac{1-\sqrt{1-x}}{x}\tag{1} $$
hence
$$ \frac{f(x)}{1-x}=\sum_{i\geq 0}(A_0+\ldots+A_i)x^i = 2 \frac{1-\sqrt{1-x}}{x(1-x)} \tag{2} $$
and the wanted sum is the coefficient of $x^{n-1}$ in the RHS of $(2)$, or the coefficient of $x^n$ in 
$$ \frac{2}{1-x}-\frac{2}{\sqrt{1-x}}\tag{3} $$
whose Taylor series is well-known and leads to
$$ \sum_{i=0}^{n-1}A_i = 2-\frac{2}{4^n}\binom{2n}{n}\tag{4} $$
as wanted.
</s>