Have I discovered another property/definition of e?

Question

I may have stumbled upon a property of Euler's constant, $e$.

Conjecture: $$ \lim_{a \to 0}\prod_{n=0}^\infty (1 + a(1-a)^{n}) = e $$

Take the geometric sequence 0.5, 0.25, 0.125 ... where Tn = 0.5 * 0.5^n. Now define a new series, whose terms are equal to (1+Tn). That is: 1.5, 1.25, 1.125 ... Multiply the first $x$ terms of this sequence together, to yield $y$. As $x$ approaches infinity, $y$ approaches approximately 2.384.

The geometric sequence which we started with has a sum which approaches 1. Infinite other geometric sequences also sum to approach 1. A geometric sequence can be defined by two variables: it's starting term, $a$, and the ratio, $r$, between each term and the previous one. The sum of a geometric sequence approaches $a$/(1-$r$). Thus, the set of sequences whose sum approaches one are defined by the relation $a$ = 1-$r$.

My conjecture is this: as $a$ approaches 0, the infinite product of (1+Tn) approaches e. Using a spreadsheet, I have calculated the product of the first 10,000 terms when $a$ = 0.001. This yeilds approximately 2.7176, slightly less than e.

I'd appreciate if anybody could either a) show that this method does not yield $e$; or b) prove that it is $e$

Answer 1

$\sum_{n=0}^\infty \ln(1 + a(1-a)^{n})\leq\sum_{n=0}^\infty a(1-a)^{n}=1$ is immediate.

Now for a lower bound. Note that $a(1-a)\rightarrow 0$ and that $a(1-a)^n>a(1-a)^{n+1}$.

Since $\ln(1+x)$ is concave, we have $\ln(1+x)\geq \dfrac{\ln(1+d)}{d}x$ whenever $x\in[0,d]$.

Therefore, an appropriate lower bound for the sum is $\dfrac{\ln(1+d)}{d}$, where $d=a(1-a)$. Now make $a$ arbitrarily small.

Answer 2

Partial answer:

As I've added to your question, you're trying to show that $$ \lim_{a \to 0^+}\prod_{n=0}^\infty (1 + a(1-a)^{n}) = e $$ It's equivalent to show that $$ \lim_{a\to 0^+}\log\left(\prod_{n=0}^\infty (1 + a(1-a)^{n}) \right) = \lim_{a\to 0^+}\sum_{n=0}^\infty \log(1 + a(1-a)^{n}) = 1 $$ To that effect, it suffices to note that for sufficiently small (positive) $a$, $$ \sum_{n=0}^\infty \log(1 + a(1-a)^{n}) = \sum_{n=0}^\infty \sum_{m=1}^\infty \frac{(-1)^{m+1}}{m}[a(1-a)^{n}]^m = \\ \sum_{m=1}^\infty\sum_{n=0}^\infty \frac{(-1)^{m+1}}{m}[a(1-a)^{n}]^m = \\ \sum_{n=0}^\infty [a(1-a)^{n}] + \sum_{m=2}^\infty\sum_{n=0}^\infty \frac{(-1)^{m+1}}{m}[a(1-a)^{n}]^m =\\ \sum_{n=0}^\infty [a(1-a)^{n}] + a\sum_{m=2}^\infty \left(a^{m-1}\frac{(-1)^{m+1}}{m}\sum_{n=0}^\infty [(1-a)^{m}]^n\right) =\\ \sum_{n=0}^\infty [a(1-a)^{n}] + a\sum_{m=2}^\infty \frac{(-1)^{m+1}}{m}\cdot \frac{a^{m-1}}{1 - (1-a)^m} $$ It suffices to show that the term on the left approaches $1$ while the term on the right approaches $0$.