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Good morning, I have a big problem solving this: $\sum_{k=1}^{\infty}\frac{k!}{(2k)!}\:$

I'm trying solving this limit with test of D'Alembert, but I have a problem solving the limit.

$\lim_{k\rightarrow\infty}\frac{(k+1)!k!}{(2k+2)!2k!}=(?)$

please, help me.

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    $\begingroup$ I don't know what limit test you were attempting to use, but it is not true that if $\lim\limits_{n\to\infty}a_n\cdot a_{n+1}=0$ then $\sum\limits_{n=1}^\infty a_n$ converges. Take for counterexample $\sum\limits_{n=1}^\infty \frac{1}{n}$. Perhaps you meant to use the ratio test as described below at which point you made the mistake of multiplying the terms instead of taking the ratio of the terms. $\endgroup$ – JMoravitz May 19 '16 at 15:07
  • $\begingroup$ Regardless, the limit you write can be seen to go to zero since $0\leq\frac{(k+1)!k!}{(2k+2)!(2k)!}\leq \frac{(2k+2)!(2k-1)!}{(2k+2)!(2k)!}=\frac{1}{2k}\to 0$ $\endgroup$ – JMoravitz May 19 '16 at 15:11
  • $\begingroup$ $\frac {k!}{(2k)!} < \frac {1}{k!}$ for all k, $\sum \frac {1}{k!} = e.$ $\endgroup$ – Doug M May 19 '16 at 17:20
  • $\begingroup$ The exact value of the sum is also calculated in these posts: math.stackexchange.com/questions/1885428/… and math.stackexchange.com/questions/2354004/… (Of course, this is much more difficult question than finding whether the series converges.) $\endgroup$ – Martin Sleziak Jul 16 '17 at 9:02
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We have that $$ \lim_{k\to\infty}\frac{(k+1)!(2k)!}{k!(2(k+1))!}=\lim_{k\to\infty}\frac{(k+1)(2k)!}{(2k+2)!}=\lim_{k\to\infty}\frac{k+1}{(2k+2)(2k+1)}=\lim_{k\to\infty}\frac1{2(2k+1)}=0. $$ Hence, the series converges by the ratio test.

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Note that

$$ \frac{(k+1)!}{(2k+2)!}\cdot \frac{(2k)!}{k!}=\frac{k+1}{(2k+2)(2k+1)}=\frac{1}{2(2k+1)}\to 0 \text{ as }k \to\infty. $$

Thus, by the ratio test, the series converges.

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$$S=\sum_{k=1}^{+\infty}\frac{k!}{(2k)!}=\sum_{k\geq 1}\frac{1}{(2k)!}\int_{0}^{+\infty} x^k e^{-x}\,dx =\int_{0}^{+\infty}e^{-x}\sum_{k\geq 1}\frac{x^k}{(2k)!}\,dx\tag{1}$$ hence: $$ S = 2\int_{0}^{+\infty} z e^{-z^2} \sum_{k\geq 1}\frac{z^{2k}}{(2k)!}\,dz = \int_{0}^{+\infty} 2z\,e^{-z^2}\left(\cosh(z)-1\right)\,dz \tag{2}$$ or: $$ S = \int_{0}^{+\infty} e^{-z^2}\sinh(z)\,dz = \color{red}{\frac{\sqrt{\pi}}{2} e^{1/4}\,\text{erf}\left(\frac{1}{2}\right)}\approx 0.5922965.\tag{3}$$

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  • $\begingroup$ Props for answering the question actually written, whether it was the intended question or not. Other answers made the tacit assumption that all that was being questioned was the convergence of the series, but not what it actually converged to. $\endgroup$ – JMoravitz May 19 '16 at 19:09
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} \color{#f00}{\sum_{k = 1}^{\infty}{k! \over \pars{2k}!}} & = \sum_{k = 1}^{\infty}{1 \over \pars{k - 1}!}\, {\Gamma\pars{k + 1}\Gamma\pars{k} \over \Gamma\pars{2k + 1}} = \sum_{k = 1}^{\infty}{1 \over \pars{k - 1}!} \int_{0}^{1}t^{k}\pars{1 - t}^{k - 1}\,\dd t \\[3mm] & = \int_{0}^{1}t\sum_{k = 0}^{\infty}{\bracks{t\pars{1 - t}}^{k} \over k!}\,\dd t = \int_{0}^{1}t\expo{t\pars{1 - t}}\,\dd t = \int_{-1/2}^{1/2}\pars{t + \half}\expo{1/4 - t^{2}}\,\dd t \\[3mm] & = {\root{\pi} \over 2}\expo{1/4}\ \overbrace{\bracks{{2 \over \root{\pi}}\int_{0}^{1/2}\expo{-t^{2}}\,\dd t}} ^{\ds{\textrm{erf}\pars{1/2}}}\ =\ \color{#f00}{{\root{\pi} \over 2}\expo{1/4}\textrm{erf}\pars{\half}} \approx 0.5923 \end{align}

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