TITLE: Show that:$\lim\limits_{x\to \infty}{\gamma\over x}+e^{-\gamma/x}\prod\limits_{y=1}^{\infty}\left({xy\over 1+xy}\right){e^{1/(xy)}}=1$
QUESTION [5 upvotes]: I always want to know if this limit hold or not.
Please show that 
$$\lim_{x\to \infty}\left[{\gamma\over x}+e^{-{\gamma\over x}}\prod_{y=1}^{\infty}\left({xy\over 1+xy}\right){e^{{1\over xy}}}\right]=1\tag1$$
Where $\gamma$ is Euler's constant; $\gamma=0.577216...$
My try: 
Re write
$$\left(1-{\gamma\over x}\right)e^{\gamma \over x}=\prod_{y=1}^{\infty}\left({xy\over 1+xy}\right){e^{1\over xy}}$$
Take the ln 
$${\gamma\over x}+\ln{\left(1-{\gamma\over x}\right)}=\sum_{y=1}^{\infty}\ln{\left(xy\over 1+xy\right){e^{1\over xy}}}$$
$${\gamma\over x}+\ln{\left(1-{\gamma\over x}\right)}=\ln{\left({x\over 1+x}\right){e^{1\over x}}}+\ln{\left({2x\over 1+2x}\right){e^{1\over 2x}}}+
\ln{\left({3x\over 1+3x}\right){e^{1\over 3x}}}+\cdots$$
any hints what to do next? Or else prove $(1)$

REPLY [3 votes]: THIS IS A RESPONSE TO THE ORIGINALLY POSTED QUESTION
Note that we have
$$\begin{align}
\log\left(\left(\frac{xy}{1+xy}\right)^{e^{\frac1{xy}}}\right)&={e^{\frac1{xy}}\log\left(1-\frac{1}{1+xy}\right)}\\\\
&=\left(1+\frac1{xy}+O\left(\frac{1}{(xy)^2}\right)\right)\left(-\frac{1}{1+xy}+O\left(\frac{1}{(1+xy)^2}\right)\right)\\\\
&=-\frac1{xy}+O\left(\frac1{(xy)^2}\right)
\end{align}$$
Inasmuch as $\lim_{N\to \infty}\sum_{y=1}^N \frac{-1}{xy}=-\infty$, the infinite product approaches $0$.

Note that if instead of the term $\left(\frac{xy}{1+xy}\right)^{e^{\frac{1}{xy}}}$ we had $\left(\frac{xy}{1+xy}\right)\,e^{\frac{1}{xy}}$, then the product represents 
$$e^{-\gamma/x}\prod_{y=1}^\infty \left(\frac{xy}{1+xy}\right)\,e^{\frac{1}{xy}}=\frac1x \Gamma(1/x)\to 1\,\,\text{as}\,\,x\to \infty$$