How find this value $\prod_{k=1}^{\infty}\left(1+\frac{1}{k^5}\right)$ 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
 A: 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
A: $$\prod_{k=1}^\infty\bigg\{1+\bigg(\frac{x}{k}\bigg)^{2n+1}\bigg\}=\frac1{x^{2n+1}\cdot\displaystyle\prod_{k=0}^{2n}\Gamma\bigg(x\cdot\exp\bigg(\frac{2\pi ik}{2n+1}\bigg)\bigg)}$$
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}

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 $\ds{\pars{~\mbox{see expression}\ \eqref{1}~}}$

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