Conjecture $\lim\limits_{x\to 0}\prod\limits_{n=1}^{\infty}\left(\arctan\left(x^{n}\right)+1\right)^{\frac{1}{x}}=e$

Question

I cannot figure why but let me propose it:

$$\lim_{x \to 0}\prod_{n=1}^{\infty}\left(\arctan\left(x^{n}\right)+1\right)^{\frac{1}{x}}\overset?=e$$

We have a more obvious statement:

$$\lim_{x\to 0}\prod_{n=1}^{\infty}\left(\tanh\left(x^{n}\right)+1\right)^{\frac{1}{x}}=e$$

If we replace a $\arctan$ by a similar function, for example $\operatorname{erf}$, we don't have the same result.

If the conjecture is true I think we may need complex numbers to prove it but it's my supposition.

I have checked the conjecture up to $n=100$ with Desmos. How should I prove or disprove it?

Answer 1

Using $O$ notation and $\arctan(x)=x+O(x^3)$ you have $$\log\left(\prod_{n=1}^{\infty}\left(\arctan\left(x^{n}\right)+1\right)^{\frac{1}{x}}\right)=\frac{1}{x}\sum_{n=1}^\infty \log\left(1+x^n+O(x^{3n})\right) = \frac1x \sum_{n=1}^\infty \left(x^n + O(x^{2n}) \right)\\ =\frac1x \left( \frac{x}{1-x} + O\left(\frac{x^2}{1-x^2}\right)\right)=\frac{1}{1-x}+O(x) = 1 + O(x) \, .$$ Your statement is true for any function $f$ with $f(x)=x+o(x)$ about $x=0$.

Answer 2

Let us denote $S(x) = \sum \limits_{n=2}^{\infty} \log\big(\arctan(x^n)+1\big)$, for $x \in \big[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\big]$ (so it is well defined; see later for the proof of the proof that it converges). We want to show that $S(x) = O(x^2)$.

For $x \ge 0$ the bound is very simple to write. Let us directly write the general case where $x$ can also be negative. We will use the two following inequalities:

• for all $x \in \big[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\big]$ $n \ge 2$, $|\arctan(x^n)| \le |x|^n \le \frac{1}{2}$,

• for $t \in \big[-\frac{1}{2}, \frac{1}{2}\big]$, $|\log(1+t)| \le |t|+t^2$. A proof can be found here for $-\frac{1}{2} \le t \le 0$, and the result is obvious for positive $t$.

Now we write, for $x \in \big[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\big]$:

\begin{align*} |S(x)| & \le \sum \limits_{n=2}^{\infty} \big|\log\big(\arctan(x^n)+1\big)\big| \\ & \le \sum \limits_{n=2}^{\infty} \big|\arctan(x^n)\big| + \arctan(x^n)^2 \\ & \le \sum \limits_{n=2}^{\infty} |x|^n + |x|^{2n} = \frac{|x|^2}{1-|x|} + \frac{x^4}{1-x^2} \end{align*}

and thus $S(x)=O(x^2)$; and this also proves that the series converges for these values of $x$. This allows us to conclude for your initial question, since for these $x$, the series converges, so the infinite product also does and we can write:

\begin{align*} \prod \limits_{n=1}^\infty \big(\arctan(x^n)+1\big)^{\frac{1}{x}} & = \exp\Big(\frac{\log\big(1+\arctan(x)\big)}{x} + \frac{1}{x}\sum\limits_{n=2}^{\infty} \log\big(1+\arctan(x^n)\big)\Big) \\ & = \exp\big(1 + O(x)\big) \end{align*}

since $\arctan(x) = x+O(x^3)$ and $\log(1+x)=x+O(x^3)$. The limit is indeed $e$.

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Erratum on my initial comment about this also working for $\mbox{erf}$: I didn't realize that the derivative at $0$ was actually important for the actual value of the limit. If you replace $\arctan$ with a function that has a derivative $\alpha$ at $0$, a similar product will converge to $\exp(\alpha)$. Again, that would be because of the first term of the series, the rest of the terms being negligible.