<s>[INST] Strange behavior of infinite products $\prod^{\infty}_{n=1} \ln (1+ \frac{1}{n} )^n$ and $\prod^{\infty}_{n=1} \ln (1+ \frac{1}{n} )^{n+1}$ There are two expressions marking the lower and upper bounds for number $e$:
$$\left(1+\frac{1}{n} \right)^n \leq e \leq \left(1+\frac{1}{n} \right)^{n+1}$$
Naturally, I wanted to know if infinite products of their logarithms converge to the same value. I was greatly surprised to find that not only do they not converge to the same value, but one of them converges to zero and the other to infinity:
$$\prod^{\infty}_{n=1} \ln \left(1+ \frac{1}{n} \right)^n=0$$ 
$$\prod^{\infty}_{n=1} \ln \left(1+ \frac{1}{n} \right)^{n+1} \rightarrow + \infty$$
On the other hand their product (or equally, the infinite product of their geometric means) converges, but not to $1$:
$$\prod^{\infty}_{n=1} \ln \left(1+ \frac{1}{n} \right)^{n} \ln \left(1+ \frac{1}{n} \right)^{n+1}=\prod^{\infty}_{n=1} n(n+1) \ln^2 \left(1+ \frac{1}{n} \right) \rightarrow P$$
Mathematica gives the following values (since $P_n$ is decreasing, it's certainly less than $1$):
$$P(14999)=0.921971686261$$
$$P(15000)=0.921971685920$$
The convergence (or divergence) can be proved using the corresponding series and the integral test:
$$\sum^{\infty}_{n=1} \ln \ln \left(1+ \frac{1}{n} \right)^n $$
$$\int^{\infty}_{1} \ln \left( x \ln \left(1+ \frac{1}{x} \right)\right) dx =\int^{1}_{0} \frac{1}{y^2} \ln  \left( \frac{\ln \left(1+ y \right)}{y} \right) dy\rightarrow - \infty $$
This integral does not converge (according to Wolframalpha)
$$\sum^{\infty}_{n=1} \ln \ln \left(1+ \frac{1}{n} \right)^{n+1} $$
$$\int^{\infty}_{1} \ln \left( (x+1) \ln \left(1+ \frac{1}{x} \right)\right) dx =\int^{1}_{0} \frac{1}{y^2} \ln \left( \left(1+ \frac{1}{y} \right) \ln \left(1+ y \right) \right) dy\rightarrow + \infty $$
This integral also does not converge (according to Wolframalpha)
Finally, the 'mean' infinite product gives (see Wolframalpha):
$$\int^{1}_{0} \frac{1}{y^2} \ln \left( \frac{1}{y} \left(1+ \frac{1}{y} \right) \ln^2 \left(1+ y \right) \right) dy=-0.0569274$$
So, this infinite product converges, but not to $1$ according to Mathematica.

Is there any explanation for all this? Is it connected to the special 
  properties of $e$?

 [/INST] Note $$\log\left(1+\frac 1n\right)=\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac1{n^2}\right)$$
So $$\log\left(1+\frac 1n\right)^n=n\log\left(1+\frac 1n\right)= 1-\frac{1}{2n}+o\left(\frac1n\right).$$
Now, since $\prod \left(1-\frac{1}{2n}\right)$ does not converge to a positive value, neither does the left side product.
You can similarly deduce:
$$(n+1)\log\left(1+\frac 1n\right) =1+\frac{1}{2n}+o\left(\frac1n\right).$$
So this is not so strange.
For the product, you need more terms. You can show:
$$\log\left(1+\frac1n\right)^2 = \frac1{n^2} -\frac{1}{n^3} + \frac{11}{12}\frac{1}{n^4}+O\left(\frac1{n^5}\right)$$
So $$n(n+1)\log\left(1+\frac1n\right)^2 =1-\frac{1}{12n^2}+O\left(\frac{1}{n^3}\right)$$
You might be able to put tighter bounds on this to get that all the terms are less than $1$. This only shows that all but finitely many are less than $1$.
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