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The sum of series

$\displaystyle 1+\frac{1}{1!}\cdot \frac{1}{4}+\frac{1\cdot 3}{2!}\cdot \frac{1}{4^2}+\frac{1\cdot 3 \cdot 5}{3!}\cdot \frac{1}{4^3}+\cdots $

what i try:

$$1+\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{1\cdot 3\cdot \cdots (2r-1)}{r!\cdot 4^r}$$

$$1+\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{(2r)!}{r!\prod^{n}_{r=1}(2r)!}$$

How do i solve it Help me please

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    $\begingroup$ what is the question...? $\endgroup$ – Jneven Mar 23 at 12:49
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The series under consideration is

\begin{align*} \color{blue}{1+\sum_{r=1}^\infty\frac{(2r-1)!!}{r!4^r}}&=1+\sum_{r=1}^\infty\frac{(2r)!}{r!4^r(2r)!!}\\ &=1+\sum_{r=1}^\infty\frac{(2r)!}{r!r!}\left(\frac{1}{8}\right)^r\\ &=\sum_{r=0}^\infty\binom{2r}{r}\left(\frac{1}{8}\right)^r\\ &=\left.\frac{1}{\sqrt{1-4z}}\right|_{z=1/8}\tag{1}\\ &=\frac{1}{\sqrt{1-\frac{1}{2}}}\\ &\,\,\color{blue}{=\sqrt{2}} \end{align*}

In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{1 + {1 \over 1!}\,{1 \over 4} + {1\cdot 3 \over 2!}\, {1 \over 4^{2}} + {1\cdot 3 \cdot 5 \over 3!}\,{1 \over 4^{3}} + \cdots} = 1 + \sum_{n = 1}^{\infty}{\prod_{k = 0}^{n - 1}\pars{2k + 1} \over n!\, 4^{n}} \\[5mm] = &\ 1 + \sum_{n = 1}^{\infty}{2^{n}\prod_{k = 0}^{n - 1}\pars{k + 1/2} \over n!\, 4^{n}} = 1 + \sum_{n = 1}^{\infty}{\pars{1/2}^{\overline{n}} \over n!} \,\pars{1 \over 2}^{n} \\[5mm] = &\ 1 + \sum_{n = 1}^{\infty}{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2} \over n!} \,\pars{1 \over 2}^{n} = \sum_{n = 0}^{\infty}{\pars{n - 1/2}! \over n!\pars{-1/2}!} \,\pars{1 \over 2}^{n} \\[5mm] = &\ \sum_{n = 0}^{\infty}{n - 1/2 \choose n}\,\pars{1 \over 2}^{n} = \sum_{n = 0}^{\infty}{-1/2 \choose n}\,\pars{-\,{1 \over 2}}^{n} = \bracks{1 + \pars{-1 \over 2}}^{-1/2} \\[5mm] = &\ \bbx{\root{2}} \approx 1.4142 \end{align}

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