How find this value $\prod_{k=1}^{\infty}\left(1+\frac{1}{k^5}\right)$

Question

Find the value $$\prod_{k=1}^{\infty}\left(1+\dfrac{1}{k^5}\right)$$

I know this :How find this $\prod_{n=2}^{\infty}\left(1-\frac{1}{n^6}\right)$

and maybe can find the $2k+1$? can you someone konw somepaper Research on this problem? if this have,can you link this paper? Thank you

Answer 1

We remark the following simple lemma. (This follows from the Stirling's formula, as you can check here.)

Lemma. If $\alpha_{1} + \cdots + \alpha_{n} = \beta_{1} + \cdots + \beta_{n}$, then $$ \lim_{N\to\infty} \frac{\Gamma(N+\alpha_{1}) \cdots \Gamma(N+\alpha_{n})}{\Gamma(N+\beta_{1}) \cdots \Gamma(N+\beta_{n})} = 1.$$

Now assume $p(z) = (z - \alpha_{1}) \cdots (z - \alpha_{n})$ be such that $\alpha_{1} + \cdots + \alpha_{n} = 0$. Then

$$ \prod_{k=m}^{N-1} \frac{p(k)}{k^{n}} = \prod_{k=m}^{N-1} \prod_{j=1}^{n} \frac{k - \alpha_{j}}{k} = \prod_{j=1}^{n} \frac{\Gamma(N - \alpha_{j})\Gamma(m)}{\Gamma(N)\Gamma(m-\alpha_{j})}. $$

Therefore taking $N \to \infty$, we get

$$ \prod_{k=m}^{\infty} \frac{p(k)}{k^{n}} = \frac{(m-1)!^{n}}{\Gamma(m-\alpha_{1})\cdots\Gamma(m-\alpha_{n})}. $$

Depending on situation, you may simplify it further using the Euler's reflection formula.

Answer 2

The specific result is given by \begin{align} \prod_{k=1}^{\infty} \left( 1 + \frac{1}{k^{5}} \right) = \frac{ 1}{|\Gamma(e^{2\pi i/5}) \Gamma(e^{6\pi i/5})|^{2}}. \end{align} See http://mathworld.wolfram.com/InfiniteProduct.html

Answer 3

$$\prod_{k=1}^\infty\bigg\{1+\bigg(\frac{x}{k}\bigg)^{2n+1}\bigg\}=\frac1{x^{2n+1}\cdot\displaystyle\prod_{k=0}^{2n}\Gamma\bigg(x\cdot\exp\bigg(\frac{2\pi ik}{2n+1}\bigg)\bigg)}$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \prod_{k = 1}^{N}\pars{1 + {1 \over k^{5}}} & = \prod_{k = 1}^{N}\pars{k^{5} + 1 \over k^{5}} = {\prod_{k = 1}^{N}\pars{k - \bar{r}_{2}}\pars{k - \bar{r}_{1}}\pars{k + 1} \pars{k - r_{1}}\pars{k - r_{2}} \over \pars{N!}^{5}} \end{align} where $\ds{r_{1} = \exp\pars{{\pi \over 5}\,\ic}}$ and $\ds{r_{2} = \exp\pars{{3\pi \over 5}\,\ic}}$. \begin{align} \prod_{k = 1}^{N}\pars{1 + {1 \over k^{5}}} & = {\pars{N + 1}! \over \pars{N!}^{5}}\, \verts{\prod_{k = 1}^{N}\pars{k - \bar{r}_{2}}}^{\,2} \verts{\prod_{k = 1}^{N}\pars{k - \bar{r}_{1}}}^{\,2} \\[5mm] & = {N + 1 \over \pars{N!}^{4}}\, \verts{\pars{1 - r_{2}}^{\overline{N}}}^{\,2} \verts{\pars{1 - r_{1}}^{\overline{N}}}^{\,2} \\[5mm] & = {N + 1 \over \verts{\Gamma\pars{1 - r_{1}}}^{\,2}\verts{\Gamma\pars{1 - r_{2}}}^{\,2}}\, \verts{\pars{N - r_{2}}! \over N!}^{\,2}\verts{\pars{N - r_{1}}! \over N!}^{\,2} \label{1}\tag{1} \end{align}

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Moreover, \begin{align} \verts{\pars{N - r}! \over N!}^{\,2} & \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, \verts{\root{2\pi}\pars{N - r}^{N - r + 1/2}\expo{-\pars{N - r}} \over \root{2\pi}N^{N + 1/2}\expo{-N}}^{\,2} = \verts{N^{N - r + 1/2}\pars{1 - r/N}^{N - r + 1/2}\expo{r} \over N^{N + 1/2}}^{\,2} \\[5mm] & \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, \verts{N^{-r}}^{\,2} = \verts{\exp\pars{-r\ln\pars{N}}}^{\,2} = \exp\pars{-2\,\Re\pars{r}\ln\pars{N}} = {1 \over N^{2\,\Re\pars{r}}} \end{align}

Note that $\ds{{1 \over N^{2\,\Re\pars{r_{1}}}}\,{1 \over N^{2\,\Re\pars{r_{2}}}} = {1 \over N}}$ such that $\ds{\pars{~\mbox{see expression}\ \eqref{1}~}}$

$$ \bbx{\prod_{k = 1}^{\infty}\pars{1 + {1 \over k^{5}}} = {1 \over \verts{\Gamma\pars{1 - r_{1}}}^{\,2}\verts{\Gamma\pars{1 - r_{2}}}^{\,2}}} \,,\qquad \left\{\begin{array}{rcl} \ds{r_{1}} & \ds{=} & \ds{\exp\pars{{\pi \over 5}\,\ic}} \\[2mm] \ds{r_{2}} & \ds{=} & \ds{\exp\pars{{3\pi \over 5}\,\ic}} \end{array}\right. $$