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.
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{i = 0}^{n - 1}{2i \choose i}{1 \over i + 1}\,{1 \over 2^{2i}} & =
\bracks{z^{n}}\sum_{\ell = 1}^{\infty}z^{\ell}\bracks{\sum_{i = 0}^{\ell - 1}{2i \choose i}{1 \over i + 1}\,{1 \over 2^{2i}}} =
\bracks{z^{n}}\sum_{i = 0}^{\infty}{2i \choose i}{1 \over i + 1}
\,{1 \over 2^{2i}}\sum_{\ell = i + 1}^{\infty}z^{\ell}
\\[5mm] & =
\bracks{z^{n}}\sum_{i = 0}^{\infty}{2i \choose i}{1 \over i + 1}
\,{1 \over 2^{2i}}\sum_{\ell = 0}^{\infty}z^{\ell + i + 1}
\\[5mm] & =
\bracks{z^{n - 1}}\pars{1 - z}^{-1}
\sum_{i = 0}^{\infty}\
\overbrace{{-1/2 \choose i}\pars{-4}^{i}}^{\ds{2i \choose i}}\
\pars{z \over 4}^{i}\int_{0}^{1}x^{i}\,\dd x
\\[5mm] & =
\bracks{z^{n - 1}}\pars{1 - z}^{-1}
\int_{0}^{1}\sum_{i = 0}^{\infty}{-1/2 \choose i}\pars{-zx}^{i}\,\dd x
\\[5mm] & =
\bracks{z^{n - 1}}\pars{1 - z}^{-1}\int_{0}^{1}\pars{1 - zx}^{-1/2}\,\dd x
\\[5mm] & =
\bracks{z^{n - 1}}\pars{1 - z}^{-1}\,{2 - 2\pars{1 - z}^{1/2} \over z}
\\[5mm] & =
2\bracks{z^{n}}\pars{1 - z}^{-1} -
2\bracks{z^{n}}\pars{1 - z}^{-1/2} =
2 - 2{-1/2 \choose n}\pars{-1}^{n}
\\[5mm] & =
2 - 2\,{{2n \choose n} \over \pars{-4}^{n}}\,\pars{-1}^{n} =
\bbx{\ds{2\bracks{1 - {1 \over 2^{2n}}{2n \choose n}}}}
\end{align}

See a
  $\ds{{2i \choose i} = {-1/2 \choose i}\pars{-4}^{i}}$ proof in
  this link.

