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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 $(1)$

  • 0
    Wow. Just... wow. – 2017-01-02
  • 1
    Have you encountered the product expansion of the Gamma function? – 2017-01-02
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    what evidence do we have that the limit holds? – 2017-01-02
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    The limit should be 0, not 1, I suspect... – 2017-01-02
  • 0
    @tired As written, the product diverges. – 2017-01-02
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    @Dr.MV thanks for the clarification (+1) – 2017-01-02
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    @tired Happy New Year my friend! – 2017-01-02
  • 0
    @Dr.MV a happy new year for you too, of course! – 2017-01-02

1 Answers 1

3

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

$$e^{-\gamma/x}\prod_{y=1}^\infty \left(\frac{xy}{1+xy}\right)\,e^{\frac{1}{xy}}=\frac1x \Gamma(1/x)\to 1\,\,\text{as}\,\,x\to \infty$$

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    And this also shows, that at least the sequence of the partial products converges to 0. Nice :) – 2017-01-02
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    You are right @Dr. MV I wrote it wrong. I will edit it. (+1) should be as you said $\left({xy\over 1+xy}\right)e^{1\over xy}$ – 2017-01-02