Evaluate the following product
$$P =\left({\frac{2}{1}}\right)^{1/8}\cdot\left({\frac{{2\cdot 2}}{{1\cdot 3}}}\right)^{3/16}\cdot\left({\frac{{2\cdot 2\cdot 2\cdot 4}}{{1\cdot 3\cdot 3\cdot 3}}}\right)^{6/32}\cdot\left({\frac{{2\cdot 2\cdot 2\cdot 2\cdot 4\cdot 4\cdot 4\cdot 4}}{{1\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 5}}}\right)^{10/64}\cdots$$
Solution
Taking logarithms at both sides, it becomes clear that:
$$\log P = - \sum_{n=1}^{\infty} \frac{n(n+1)}{2^{n+3}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \log(k+1)$$
However, the right hand side is:
\begin{align*}
\sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \log(k+1) &= \lim_{s\to 0^{+}} \Gamma(s+1) \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \frac{(k+1)^{-s} - 1}{(-s)} \\ &= - \lim_{s\to 0^{+}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \frac{\Gamma(s)}{(k+1)^{s}} \\ &= - \lim_{s\to 0^{+}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \int_{0}^{\infty} x^{s-1} e^{-(k+1)x} \, {\rm }x \\ &= - \lim_{s\to 0^{+}} \int_{0}^{\infty} x^{s-1} e^{-x} (1 - e^{-x})^{n} \, {\rm d}x \\ &= - \int_{0}^{\infty} e^{-x} (1 - e^{-x})^{n} \, \frac{{\rm d} x}{x}
\end{align*}
This in return gives
\begin{align*}\log P &= \sum_{n=1}^{\infty} \frac{n(n+1)}{2^{n+3}} \int_{0}^{\infty} e^{-x} (1 - e^{-x})^{n} \, \frac{{\rm d}x}{x} \\ &= \frac{1}{8} \int_{0}^{\infty} e^{-x} \sum_{n=1}^{\infty} n(n+1) \left( \frac{1 - e^{-x}}{2} \right)^{n} \, \frac{{\rm d}x}{x} \\ &= \int_{0}^{\infty} x^{-1} \frac{e^{-x}(1 - e^{-x})}{(1+e^{-x})^{3}} \, {\rm d}x \end{align*}
Let us introduce now the function
$$\mathcal{J}(s)= \int_{0}^{\infty} x^{s-1} \frac{e^{-x}(1 - e^{-x})}{(1+e^{-x})^{3}} \,{\rm d}x$$
We note that $\mathcal{J}(s)$ is analytic for $\mathfrak{Re}(s)>-1$ and (fair and square) $\log P= \lim \limits_{s \rightarrow 0+} \mathcal{J}(s)$. However,
\begin{align*} \mathcal{J}(s) &= \int_{0}^{\infty} x^{s-1} \frac{e^{-x}(1 - e^{-x})}{(1+e^{-x})^{3}} \, {\rm d}x \\ &= \int_{0}^{\infty} x^{s-1} \sum_{n=1}^{\infty} (-1)^{n-1} n^2 e^{-n x} \, {\rm d}x \\ &= \sum_{n=1}^{\infty} (-1)^{n-1} n^2 \int_{0}^{\infty} x^{s-1} e^{-n x} \, {\rm d}x \\ &= \Gamma(s) \eta(s-2) \\&= \Gamma(s) (1 - 2^{3-s}) \zeta(s-2) \end{align*}
The result remains valid on any open subset of $\mathbb{C}$ where $\mathcal{J}(s)$ is analytic. Taking limit as $s \rightarrow 0+$ we get that $ \log P = -7 \zeta'(-2)$. Differentiating the functional equation of the $\zeta$ function , that is:
$$\zeta(s) = 2^{s} \pi^{s-1} \sin \left( \frac{\pi s}{2} \right) \Gamma (1-s) \zeta(1-s)$$
and plugging $s=-2$ we get that $ \displaystyle \zeta'(-2) = -\frac{\zeta(3)}{4\pi^2} = -\frac{\zeta(3)}{24\zeta(2)}$. The result follows.
$$P =\left({\frac{2}{1}}\right)^{1/8}\cdot\left({\frac{{2\cdot 2}}{{1\cdot 3}}}\right)^{3/16}\cdot\left({\frac{{2\cdot 2\cdot 2\cdot 4}}{{1\cdot 3\cdot 3\cdot 3}}}\right)^{6/32}\cdot\left({\frac{{2\cdot 2\cdot 2\cdot 2\cdot 4\cdot 4\cdot 4\cdot 4}}{{1\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 5}}}\right)^{10/64}\cdots$$
Solution
Taking logarithms at both sides, it becomes clear that:
$$\log P = - \sum_{n=1}^{\infty} \frac{n(n+1)}{2^{n+3}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \log(k+1)$$
However, the right hand side is:
\begin{align*}
\sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \log(k+1) &= \lim_{s\to 0^{+}} \Gamma(s+1) \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \frac{(k+1)^{-s} - 1}{(-s)} \\ &= - \lim_{s\to 0^{+}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \frac{\Gamma(s)}{(k+1)^{s}} \\ &= - \lim_{s\to 0^{+}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \int_{0}^{\infty} x^{s-1} e^{-(k+1)x} \, {\rm }x \\ &= - \lim_{s\to 0^{+}} \int_{0}^{\infty} x^{s-1} e^{-x} (1 - e^{-x})^{n} \, {\rm d}x \\ &= - \int_{0}^{\infty} e^{-x} (1 - e^{-x})^{n} \, \frac{{\rm d} x}{x}
\end{align*}
This in return gives
\begin{align*}\log P &= \sum_{n=1}^{\infty} \frac{n(n+1)}{2^{n+3}} \int_{0}^{\infty} e^{-x} (1 - e^{-x})^{n} \, \frac{{\rm d}x}{x} \\ &= \frac{1}{8} \int_{0}^{\infty} e^{-x} \sum_{n=1}^{\infty} n(n+1) \left( \frac{1 - e^{-x}}{2} \right)^{n} \, \frac{{\rm d}x}{x} \\ &= \int_{0}^{\infty} x^{-1} \frac{e^{-x}(1 - e^{-x})}{(1+e^{-x})^{3}} \, {\rm d}x \end{align*}
Let us introduce now the function
$$\mathcal{J}(s)= \int_{0}^{\infty} x^{s-1} \frac{e^{-x}(1 - e^{-x})}{(1+e^{-x})^{3}} \,{\rm d}x$$
We note that $\mathcal{J}(s)$ is analytic for $\mathfrak{Re}(s)>-1$ and (fair and square) $\log P= \lim \limits_{s \rightarrow 0+} \mathcal{J}(s)$. However,
\begin{align*} \mathcal{J}(s) &= \int_{0}^{\infty} x^{s-1} \frac{e^{-x}(1 - e^{-x})}{(1+e^{-x})^{3}} \, {\rm d}x \\ &= \int_{0}^{\infty} x^{s-1} \sum_{n=1}^{\infty} (-1)^{n-1} n^2 e^{-n x} \, {\rm d}x \\ &= \sum_{n=1}^{\infty} (-1)^{n-1} n^2 \int_{0}^{\infty} x^{s-1} e^{-n x} \, {\rm d}x \\ &= \Gamma(s) \eta(s-2) \\&= \Gamma(s) (1 - 2^{3-s}) \zeta(s-2) \end{align*}
The result remains valid on any open subset of $\mathbb{C}$ where $\mathcal{J}(s)$ is analytic. Taking limit as $s \rightarrow 0+$ we get that $ \log P = -7 \zeta'(-2)$. Differentiating the functional equation of the $\zeta$ function , that is:
$$\zeta(s) = 2^{s} \pi^{s-1} \sin \left( \frac{\pi s}{2} \right) \Gamma (1-s) \zeta(1-s)$$
and plugging $s=-2$ we get that $ \displaystyle \zeta'(-2) = -\frac{\zeta(3)}{4\pi^2} = -\frac{\zeta(3)}{24\zeta(2)}$. The result follows.
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