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
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.
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
$$\prod_{k=1}^\infty\bigg\{1+\bigg(\frac{x}{k}\bigg)^{2n+1}\bigg\}=\frac1{x^{2n+1}\cdot\displaystyle\prod_{k=0}^{2n}\Gamma\bigg(x\cdot\exp\bigg(\frac{2\pi ik}{2n+1}\bigg)\bigg)}$$
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
Note that $\ds{{1 \over N^{2\,\Re\pars{r_{1}}}}\,{1 \over N^{2\,\Re\pars{r_{2}}}} = {1 \over N}}$ such that $\ds{\pars{~\mbox{see expression}\ \eqref{1}~}}$
$$ \bbx{\prod_{k = 1}^{\infty}\pars{1 + {1 \over k^{5}}} = {1 \over \verts{\Gamma\pars{1 - r_{1}}}^{\,2}\verts{\Gamma\pars{1 - r_{2}}}^{\,2}}} \,,\qquad \left\{\begin{array}{rcl} \ds{r_{1}} & \ds{=} & \ds{\exp\pars{{\pi \over 5}\,\ic}} \\[2mm] \ds{r_{2}} & \ds{=} & \ds{\exp\pars{{3\pi \over 5}\,\ic}} \end{array}\right. $$
