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 $e$?

  • $\begingroup$ Try to make a series expansion of $\ln(1+1/n)^n$ and $\ln(1+1/n)^{n+1}$ (and of their product) for large $n$. The result you see can be explain from the form of the leading terms of these series. Some useful facts: If $a_n > 0$ then $\prod (1+a_n)$ converges if and only if $\sum a_n$ converges. If $1 > a_n > 0$ then $\prod (1-a_n)$ converges to $0$ if $\sum a_n$ diverges. $\endgroup$ – Winther Feb 12 '16 at 17:28

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 $1$.

  • $\begingroup$ This also explains why their product is slightly less than 1. Thank you $\endgroup$ – Yuriy S Feb 12 '16 at 17:45
  • 1
    $\begingroup$ It doesn't quite work for the product, because, if you elaborate further and use $O(1/n^2)$ where I've used $o(1/n)$, then you get that the product is $1-\frac{1}{4n^2} +O(1/n^2)=1+O(1/n^2)$, which could be greater than or less than $1$, because the $O(1/n^2)$ term "swallows" the $\frac{1}{4n^2}$ term. $\endgroup$ – Thomas Andrews Feb 12 '16 at 17:49

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