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.
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What do you know about Catalan numbers and their interpretation as paths in a grid? – Jack D'Aurizio Mar 23 '17 at 07:17
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Is it $\binom{2n}{i}$ in LHS? – Jaideep Khare Mar 23 '17 at 07:17
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I have no idea about catalan numbers and their interpretation. – J. Griez Mar 23 '17 at 07:18
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No, it is $\binom{2i}{i}$. – J. Griez Mar 23 '17 at 07:19
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1@J.Griez: in such a case, where did this problem arise? Are you confident in the use of generating functions? I am pretty sure this problem appears in Wilf - Generatingfunctionology (https://www.math.upenn.edu/~wilf/gfology2.pdf) – Jack D'Aurizio Mar 23 '17 at 07:42
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@JackD'Aurizio thanks for your answer. I am confident with generating functions. Could you please mention in which page Gilf mentioned that? – J. Griez Mar 23 '17 at 07:48
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@J.Griez: I was wrong, Wilf does not mention such exercise. Maybe it was mentioned by Flajolet in Analytic Combinatorics. – Jack D'Aurizio Mar 23 '17 at 07:57
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Near identical duplicate of this question asked less than 9 hours previous. The main difference is that there were more hints in the older question and there was no restriction on the method. – N. Shales Mar 23 '17 at 14:31
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Wouldn't this appear to be a duplicate, see this MSE link? – Marko Riedel Mar 23 '17 at 20:40
2 Answers
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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.
Jack D'Aurizio
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.
Felix Marin
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