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Hi my homework question is to prove $$ \sum_{n = 0}^{\infty}\frac{\left(2n\right)!}{4^{n}\left(n+1\right)!\left(n\right)!} = 2. $$

  • I know the sequence of partial sums $S_n$ converges from trying the ratio test.
  • I tried to prove that $S_n\rightarrow 2$ as $n\rightarrow \infty$ by expressing writing $\frac{1}{1-x}$ with taylor expansion so that I can represent 2 with an infinite summation and subtract that. That didn't work.
  • Next I tried to just consider each term in the summation and write $4^n=2^n2^n$ and then write $2^n(n)!=(2n)(2n-2)(2n-4)...(4)(2)$ and try to cancel terms and then write the leftover $2^n(n+1)!=(2n+2)(2n)...(4)(2))$. I was left with alternating products on the nominator and denominator, I belive it looked something like this $\frac{(2n-1)(2n-3)...(3)(1)}{(2n+2)(2n)(2n-2)...(4)(2)}$.

I don't know where to go from here and don't have other ideas to try. Please help.

Felix Marin
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Goob
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    Show that $$ \frac{{1 - \sqrt {1 - 4x} }}{{2x}} = \sum\limits_{n = 0}^\infty {\frac{1}{{n + 1}}\binom{2n}{n}x^n } $$ and let $x\to 1/4$ (refer to Abel's theorem). Take a look at the generating function of the Catalan numbers. – Gary Nov 21 '22 at 05:02
  • Ah thank you! I used this to go through lecture notes and was able to piece this equality together with some of the lecture things. That solution will be very simple haha because I will just quote the notes. – Goob Nov 21 '22 at 06:14

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The binomial theorem gives $$\frac 1{\sqrt{1-x}}=(1-x)^{-1/2} = \sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}(-1)^n x^n =\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^n$$ and hence $$\int_{0}^{z} \frac 1{\sqrt{1-x}}\;\text{d}x = \int_{0}^{z} \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^n\;\text{d}x = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\int_{0}^{z} x^n\;\text{d}x$$ which gives $$2-2\sqrt{1-z} = \sum_{n\geq 0}\frac{\binom{2n}{n}}{(n+1)4^n}\,z^{n+1}$$ Putting $z=1$, we have what you want.

Sayan Dutta
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\sum_{n = 0}^{\infty}{\pars{2n}! \over 4^{n}\pars{n + 1}!\pars{n}!}} = \sum_{n = 0}^{\infty}{1 \over 4^{n}}{\pars{2n}! \over n!\ n!}{1 \over n + 1} = \sum_{n = 0}^{\infty}{1 \over 4^{n}} \overbrace{2n \choose n}^{\ds{{-1/2 \choose n}\pars{-4}^{n}}}\overbrace{\int_{0}^{1}t^{n}\,\dd t} ^{\ds{\quad1/\pars{n + 1}}} \\[5mm] = & \ \int_{0}^{1}\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-t}^{n}\,\dd t = \int_{0}^{1}\pars{1 - t}^{-1/2}\,\,\dd t = \bbx{\color{#44f}{\large 2}} \\ & \end{align}

Felix Marin
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