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Question: Find the value of $$\lim_{n\to\infty}\sum_{k=1}^n \dfrac{1\cdot 3\cdot 5\cdots (2k-1)}{3\cdot 6\cdot 9\cdots 3k}~.$$

My work: Let $~~a_n=\dfrac{1\cdot 3\cdot 5\cdots (2k-1)}{3\cdot 6\cdot 9\cdots 3k}=\dfrac{1\cdot 2\cdot 3\cdots 2k}{3^k(1\cdot 2\cdot 3\cdots k)\cdot 2^k(1\cdot 2\cdot 3\cdots k)}=\dfrac{(2k)!}{6^k\cdot(k~!)^2}~.$

Hence our problem becomes $$L=\lim_{n\to\infty}~\sum_{k=1}^n \dfrac{1\cdot 3\cdot 5\cdots (2k-1)}{3\cdot 6\cdot 9\cdots 3k}~=\lim_{n\to\infty}~\sum_{k=1}^n\dfrac{(2k)!}{6^k\cdot(k~!)^2}=\lim_{n\to\infty}~\sum_{k=1}^n \binom {2k}k \dfrac{1}{6^k}.$$

Now we know that $~(1-4x)^{-1/2}=\sum_{k=0}^\infty \binom {2k}k x^k~$, then $$L=\lim_{n\to\infty}~\left[-1+\sum_{k=0}^n \binom {2k}k \left(\dfrac{1}{6}\right)^k\right]=-1+\lim_{n\to\infty}\sum_{k=0}^n \binom {2k}k \left(\dfrac{1}{6}\right)^{k}=-1+\left(1-\frac 46\right)^{-1/2}=~\sqrt 3 -1~.$$

Is the whole process described here is okay now ?

Special thanks to @Integrand & Z Ahmed

nmasanta
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2 Answers2

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Use $$\sum_{k=0}^{\infty} {2k \choose k} x^k=\frac{1}{\sqrt{1-4x}}$$ Hence,$$L=\lim_{n \to \infty}\sum_{k=0}^n {2k \choose k} 6^{-k}=\frac{1}{\sqrt{1-4/6}}=\sqrt{3}$$

Z Ahmed
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  • Here sum is from $~k=1\to n~$, so in continuation to your solution, the solution of the given problem is $~\sqrt 3 -1~.$ Right @ZAhmed ? – nmasanta Sep 16 '20 at 16:24
  • Yes, you are right. – Z Ahmed Sep 16 '20 at 16:32
  • Okay, thank you. Now I have edited my question according to the help of you and @Integrand. Please check the whole process and let me know if it is alright or not. – nmasanta Sep 16 '20 at 17:01
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

\begin{align} &\bbox[5px,#ffd]{\lim_{n\to\infty}\sum_{k = 1}^{n} {1\cdot 3\cdot 5\cdots \pars{2k - 1} \over 3\cdot 6\cdot 9\cdots 3k}} = \sum_{k = 1}^{\infty}{\prod_{j = 1}^{k}\pars{2j - 1} \over \prod_{\ell = 1}^{k}\pars{3\ell}} = \sum_{k = 1}^{\infty}{2^{k}\prod_{j = 1}^{k}\pars{j - 1/2} \over 3^{k}\,k!} \\[5mm] = &\ \sum_{k = 1}^{\infty}{\pars{2/3}^{k} \over k!}\, \pars{1 \over 2}^{\overline{k}} = \sum_{k = 1}^{\infty}{\pars{2/3}^{k} \over k!}\, {\Gamma\pars{1/2 + k} \over \Gamma\pars{1/2}!} = \sum_{k = 1}^{\infty} {\pars{k - 1/2}! \over k!\pars{-1/2}!}\pars{2 \over 3}^{k} \\[5mm] = &\ \sum_{k = 1}^{\infty}{k - 1/2 \choose k}\pars{2 \over 3}^{k} = \sum_{k = 1}^{\infty}{-1/2 \choose k}\pars{-\,{2 \over 3}}^{k} = \bracks{1 + \pars{-\,{2 \over 3}}}^{-1/2} - 1 \\[5mm] = &\ \bbx{\root{3} - 1} \approx 0.7321 \\ & \end{align}
Felix Marin
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