To Display the LaTeX follow this link
Problem: The limit due to Glaisher (1877) referred as (53) in wolfram page of Catalan's Contant(G):
$\displaystyle exp\left(-\frac{1}{2}+\frac{2G}{\pi}\right) = \lim\limits_{n \to \infty} \frac{1}{(4n+1)^{2n}}\prod\limits_{k=1}^{n}\frac{(4k-1)^{4k-1}}{(4k-3)^{4k-3}}$
Link: Math.Se
Solution: We start with the integral representation of Catalan's constant:
$\displaystyle G = \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} \dfrac{x}{\sin x}\,dx = \frac{\pi^2}{2}\int\limits_{0}^{\frac{1}{2}}\frac{x}{\sin \pi x}\,dx$
and we have, from partial fraction expansion of $\displaystyle \frac{\pi}{\sin x}$ ,
$$\begin{align}\frac{\pi}{\sin \pi x} &= \frac{1}{x} - 2x\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n^2 - x^2}\\
&= \sum\limits_{n=0}^{\infty} \int_0^1 (-1)^nt^{n-x}\,dt + (-1)^n t^{n-1+x} \,dt\\
&= \int_0^1 \sum\limits_{n=0}^{\infty} (-1)^nt^{n-x} \,dt + \int_0^1 \sum\limits_{n=0}^{\infty}(-1)^nt^{n-1+x} \,dt\\
&= \int_0^1 \frac{1}{t^x(1+t)}\,dt + \int_0^1 \frac{1}{t^{1-x}(1+t)} \\
&=\int_0^1 \frac{1}{t^x(1+t)}\,dt +\int_1^{\infty} \frac{1}{t^x(1+t)}\,dt \\
&= \int_0^{\infty} \frac{1}{t^x(1+t)}\,dt\end{align}$$
Then, $\displaystyle G = \frac{\pi}{2}\int\limits_{0}^{\frac{1}{2}} \int_0^{\infty} \frac{x}{t^x(1+t)}\,dt \,dx = \frac{\pi}{2}\int_0^{\infty} \int\limits_{0}^{\frac{1}{2}} \frac{x}{t^x(1+t)}\,dx\,dt$
Now, $\displaystyle \int\limits_{0}^{\frac{1}{2}} \frac{x}{t^x(1+t)}\,dx = \frac{1}{t+1}\left(\frac{1}{\log^2 t} - \frac{1}{\sqrt{t}\log^ t} - \frac{1}{2\sqrt{t}\log t} \right)$
Therefore, $\displaystyle \frac{2G}{\pi} = I = \int_0^{\infty} \frac{1}{t+1}\left(\frac{1}{\log^2 t} - \frac{1}{\sqrt{t}\log^ t} - \frac{1}{2\sqrt{t}\log t} \right)\,dt$
Making the substitiution, $t = e^{-w}$, the above integral becomes:
$$\begin{align} I &= \int_{-\infty}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(e^{-w} - e^{-w/2} + \frac{w}{2}e^{-w/2}\right)\,dw\\
&=\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(e^{-w} - e^{-w/2} + \frac{w}{2}e^{-w/2}\right)\,dw+\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(1 - e^{-w/2} - \frac{w}{2}e^{-w/2}\right)\,dw\\
&=\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(1+ e^{-w} - 2e^{-w/2}\right)\,dw\end{align}$$
Introducing regulating factor $s$, we write:
$$I(s) = \int_0^{\infty} \frac{w^{s-2}}{1+e^{-w}}\left( 1+e^{-w} -2e^{-w/2} \right)\,dw$$
Therefore, $$\begin{align} I(s) &= \sum\limits_{n=0}^{\infty} (-1)^n \int_{0}^{\infty}\left( 1+e^{-w} -2e^{-w/2} \right)w^{s-2}e^{-nw}\,dw\\
&=\sum\limits_{n=0}^{\infty} (-1)^n\left(\frac{\Gamma(s-1)}{n^{s-1}} + \frac{\Gamma(s-1)}{(n+1)^{s-1}} -\frac{2^s\Gamma(s-1)}{(2n+1)^{s-1}}\right)\end{align}$$
Using the fact that, $\displaystyle \lim\limits_{s \to 0} s\Gamma(s-1) = -1$, we have,
$$I = \lim\limits_{s \to 0} \sum\limits_{n=0}^{\infty} \frac{(-1)^{n+1}}{s}\left(\frac{1}{n^{s-1}} + \frac{1}{(n+1)^{s-1}} - \frac{2^s}{(2n+1)^{s-1}}\right)$$
Since, $$\begin{align} \lim_{s \to 0} \frac{\frac{1}{n^{s-1}} + \frac{1}{(n+1)^{s-1}} - \frac{2^s}{(2n+1)^{s-1}}}{s} &= n\left(2\log(n+1/2)-\log(n+1)-\log(n)\right) \\&+ \log(n+1/2)-\log(n+1)\end{align}$$
We have, $$\displaystyle \begin{align} I &= \sum_{n=0}^{\infty} (-1)^{n+1} n\left(2\log(n+1/2)-\log(n+1)-\log(n) \right) \\&+ \sum_{n=0}^{\infty} (-1)^{n+1} \left(\log(n+1/2)-\log(n+1)\right)\\
&= \frac{1}{2} + \lim\limits_{n \to \infty} \log \frac{1}{(4n+1)^{2n}} \prod_{k=0}^{n-1}\frac{(4k+3)^{4k+3}}{(4k+1)^{4k+1}}\end{align}$$
Thus, $$\displaystyle \exp{\frac{2G}{\pi} - \frac{1}{2}} = \lim\limits_{n \to \infty} \frac{1}{(4n+1)^{2n}} \prod\limits_{k=0}^{n-1}\frac{(4k+3)^{4k+3}}{(4k+1)^{4k+1}}$$
Problem: The limit due to Glaisher (1877) referred as (53) in wolfram page of Catalan's Contant(G):
$\displaystyle exp\left(-\frac{1}{2}+\frac{2G}{\pi}\right) = \lim\limits_{n \to \infty} \frac{1}{(4n+1)^{2n}}\prod\limits_{k=1}^{n}\frac{(4k-1)^{4k-1}}{(4k-3)^{4k-3}}$
Link: Math.Se
Solution: We start with the integral representation of Catalan's constant:
$\displaystyle G = \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} \dfrac{x}{\sin x}\,dx = \frac{\pi^2}{2}\int\limits_{0}^{\frac{1}{2}}\frac{x}{\sin \pi x}\,dx$
and we have, from partial fraction expansion of $\displaystyle \frac{\pi}{\sin x}$ ,
$$\begin{align}\frac{\pi}{\sin \pi x} &= \frac{1}{x} - 2x\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n^2 - x^2}\\
&= \sum\limits_{n=0}^{\infty} \int_0^1 (-1)^nt^{n-x}\,dt + (-1)^n t^{n-1+x} \,dt\\
&= \int_0^1 \sum\limits_{n=0}^{\infty} (-1)^nt^{n-x} \,dt + \int_0^1 \sum\limits_{n=0}^{\infty}(-1)^nt^{n-1+x} \,dt\\
&= \int_0^1 \frac{1}{t^x(1+t)}\,dt + \int_0^1 \frac{1}{t^{1-x}(1+t)} \\
&=\int_0^1 \frac{1}{t^x(1+t)}\,dt +\int_1^{\infty} \frac{1}{t^x(1+t)}\,dt \\
&= \int_0^{\infty} \frac{1}{t^x(1+t)}\,dt\end{align}$$
Then, $\displaystyle G = \frac{\pi}{2}\int\limits_{0}^{\frac{1}{2}} \int_0^{\infty} \frac{x}{t^x(1+t)}\,dt \,dx = \frac{\pi}{2}\int_0^{\infty} \int\limits_{0}^{\frac{1}{2}} \frac{x}{t^x(1+t)}\,dx\,dt$
Now, $\displaystyle \int\limits_{0}^{\frac{1}{2}} \frac{x}{t^x(1+t)}\,dx = \frac{1}{t+1}\left(\frac{1}{\log^2 t} - \frac{1}{\sqrt{t}\log^ t} - \frac{1}{2\sqrt{t}\log t} \right)$
Therefore, $\displaystyle \frac{2G}{\pi} = I = \int_0^{\infty} \frac{1}{t+1}\left(\frac{1}{\log^2 t} - \frac{1}{\sqrt{t}\log^ t} - \frac{1}{2\sqrt{t}\log t} \right)\,dt$
Making the substitiution, $t = e^{-w}$, the above integral becomes:
$$\begin{align} I &= \int_{-\infty}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(e^{-w} - e^{-w/2} + \frac{w}{2}e^{-w/2}\right)\,dw\\
&=\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(e^{-w} - e^{-w/2} + \frac{w}{2}e^{-w/2}\right)\,dw+\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(1 - e^{-w/2} - \frac{w}{2}e^{-w/2}\right)\,dw\\
&=\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(1+ e^{-w} - 2e^{-w/2}\right)\,dw\end{align}$$
Introducing regulating factor $s$, we write:
$$I(s) = \int_0^{\infty} \frac{w^{s-2}}{1+e^{-w}}\left( 1+e^{-w} -2e^{-w/2} \right)\,dw$$
Therefore, $$\begin{align} I(s) &= \sum\limits_{n=0}^{\infty} (-1)^n \int_{0}^{\infty}\left( 1+e^{-w} -2e^{-w/2} \right)w^{s-2}e^{-nw}\,dw\\
&=\sum\limits_{n=0}^{\infty} (-1)^n\left(\frac{\Gamma(s-1)}{n^{s-1}} + \frac{\Gamma(s-1)}{(n+1)^{s-1}} -\frac{2^s\Gamma(s-1)}{(2n+1)^{s-1}}\right)\end{align}$$
Using the fact that, $\displaystyle \lim\limits_{s \to 0} s\Gamma(s-1) = -1$, we have,
$$I = \lim\limits_{s \to 0} \sum\limits_{n=0}^{\infty} \frac{(-1)^{n+1}}{s}\left(\frac{1}{n^{s-1}} + \frac{1}{(n+1)^{s-1}} - \frac{2^s}{(2n+1)^{s-1}}\right)$$
Since, $$\begin{align} \lim_{s \to 0} \frac{\frac{1}{n^{s-1}} + \frac{1}{(n+1)^{s-1}} - \frac{2^s}{(2n+1)^{s-1}}}{s} &= n\left(2\log(n+1/2)-\log(n+1)-\log(n)\right) \\&+ \log(n+1/2)-\log(n+1)\end{align}$$
We have, $$\displaystyle \begin{align} I &= \sum_{n=0}^{\infty} (-1)^{n+1} n\left(2\log(n+1/2)-\log(n+1)-\log(n) \right) \\&+ \sum_{n=0}^{\infty} (-1)^{n+1} \left(\log(n+1/2)-\log(n+1)\right)\\
&= \frac{1}{2} + \lim\limits_{n \to \infty} \log \frac{1}{(4n+1)^{2n}} \prod_{k=0}^{n-1}\frac{(4k+3)^{4k+3}}{(4k+1)^{4k+1}}\end{align}$$
Thus, $$\displaystyle \exp{\frac{2G}{\pi} - \frac{1}{2}} = \lim\limits_{n \to \infty} \frac{1}{(4n+1)^{2n}} \prod\limits_{k=0}^{n-1}\frac{(4k+3)^{4k+3}}{(4k+1)^{4k+1}}$$
