Evaluate $\prod_{n=1}^{\infty}e^{\frac{(-1)^{n+1}x^n}{n}}$ for every $x\in\mathbb{R}$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be defined as:

$$f(x)=\prod_{n=1}^{\infty}e^{\frac{(-1)^{n+1}x^n}{n}}$$

Prove that $f(x)$ is well defined and continuous if and only if $x\leq1$.

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I think I did well most of what is required. I showed that:

$$f(x)=\prod_{n=1}^{\infty}e^{\frac{(-1)^{n+1}x^n}{n}}=e^{S(x)}$$

Where $S(x)\triangleq\displaystyle{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}}$.

I noticed that $S(x)$ is actually the Taylor Series of $\ln(1+x)$, and I know it converges for every $|x|<1$, according to the Ratio Test. Substituting $x=1$ I get $S(1)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$, which converges according to Leibniz Test, and for $x=-1$ I got $S(-1)=\sum_{n=1}^{\infty}-\frac{1}{n}=-\infty$.

To conclude:

$$-1<x\leq1:f(x)=e^{\ln(1+x)}=x+1\\f(-1)=e^{-\infty}=0$$

Now for every $x<-1$ I defined $y=-x>1$, which led me to:

$$S(x)=S(-y)=-\sum_{n=1}^{\infty}\frac{y^n}{n}\triangleq T(y)$$

I noticed $T(y)<0$ since $y>1$, and I know that the sum diverges. Since $T(y)$ sums up infinite number of negative numbers and diverges, I concluded that $T(y)=-\infty$, so therefore:

$$x<-1: f(x)=e^{S(x)}=e^{T(y)}=e^{-\infty}=0$$

Since $\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)=0=f(-1)$, we can conclude $f(x)$ is continuous (since $x=-1$ is the only problematic point - for every $x\neq -1$, $f(x)$ is elementary and continuous).

Now the problem is to show that $S(x)$ diverges for every $x>1$, either to $\infty$ or to nothing. I have no idea how to do that. I will say that I suspect $S(x)$ diverges to nothing. I tried using Leibniz Test, but I've always had hard time using it with function series.

Thanks!

Answer 1

Note that an infinite product of positive numbers

$$\prod_{n=1}^\infty a_n$$

converges to a nonzero real number if and only if

$$\sum_{n=1}^\infty \log(a_n)$$

converges. Check here for details. Setting

$$a_n=e^{\frac{(-1)^{n+1}x^n}{n}}$$

we check to see if

$$\sum_{n=1}^\infty \log(e^{\frac{(-1)^{n+1}x^n}{n}})=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}$$

converges for $x>1$. But then

$$\lim_{n\to\infty}\left|\frac{(-1)^{n+1}x^n}{n}\right|=\lim_{n\to\infty}\frac{x^n}{n}=\infty$$

Thus, the sum does not converge and hence the product does not converge. Finally, the infinite product can not converge to zero as

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}>x>1$$

This implies

$$\prod_{n=1}^\infty e^{\frac{(-1)^{n+1}x^n}{n}}\geq e^1=e$$

We conclude the product does not converge for $x>1$.