Left Hand Limit of this continous product

Question

Let $$\lim_{x \to 1^-} \prod_{n=0}^{\infty} \left[\frac{1+x^{n+1}}{1+x^n}\right]^{x^n}=L.$$ Find $\left\lceil 1/L \right\rceil$.

Answer 1

First, let's study the series $\Psi(x):=\sum_{n=0}^\infty x^n\log(1+x^n)$ where $x\in(0,1)$. Since $$\frac{d}{dt}\left[x^t\log(1+x^t)\right]=\dfrac{x^{2t}\log\left(x\right)}{1+x^t}+x^t\log\left(x\right)\log\left(1+x^t\right) = \underbrace{\log(x)}_{<0}\underbrace{\left(\frac{x^{2t}}{1+x^t}+x^t\log(1+x^t)\right)}_{>0}<0$$ we have that $f(t)=x^t\log(1+x^t)$ is a decreasing continuous function such that $\lim\limits_{t\to+\infty}x^t\log(1+x^t)=0$ and we can make use of the integral test to prove that $\Psi(x)$ is indeed convergent for $x\in(0,1)$. Notice that $$\int x^t\log(1+x^t)dt=\dfrac{\log\left(1+x^t\right)\left(1+x^t\right)-x^t}{\log\left(x\right)}+C$$ $$\Rightarrow\int_0^\infty x^t\log(1+x^t)dt=-\dfrac{2\log(2)-1}{\log\left(x\right)}$$ which indeed converges (the indefinite integral can be easily solved with the substitution $u=1+x^t$). Hence $\Psi(x)$ is well defined for $x\in(0,1)$. In fact, by the integral test, we also know that $$\color{blue}{-\dfrac{2\log(2)-1}{\log\left(x\right)}}=\int_{0}^\infty x^t\log(1+x^t)dt\leq\color{purple}{\Psi(x)}\leq\int_{-1}^\infty x^t\log(1+x^t)dt=\color{red}{-\dfrac{\log\left(1+x^{-1}\right)\left(1+x\right)-1}{{x\log\left(x\right)}}}$$

Now that we have already studied the said function, consider $L(x)=\prod_{n=0}^\infty\left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}$ for $x\in(0,1)$ where we wish to find $\lim\limits_{x\to1^-}L(x)$. Now, notice that $$\log\left(\prod_{n=0}^N\left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}\right)=\sum_{n=0}^Nx^n\log(1+x^{n+1})-\sum_{n=0}^Nx^n\log(1+x^{n})$$ $$=\sum_{n=1}^{N+1}x^{n-1}\log(1+x^{n})-\sum_{n=0}^Nx^n\log(1+x^{n})$$ $$=x^N\log(1+x^{N+1})-\frac{\log(2)}{x}+\sum_{n=0}^{N}x^{n-1}\log(1+x^{n})-\sum_{n=0}^Nx^n\log(1+x^{n})$$ $$=x^N\log(1+x^{N+1})-\frac{\log(2)}{x}+\left(\frac{1}{x}-1\right)\sum_{n=0}^Nx^n\log(1+x^{n})$$ which implies that, since $\lim_Nx^N\log(1+x^{N+1})=0$, the identity $\log(L(x))=-\frac{\log(2)}{x}+\left(\frac{1}{x}-1\right)\Psi(x)$ holds and $L(x)$ does indeed converge for $x\in(0,1)$. Finally, if we study the limit $\lim\limits_{x\to1^-}\left(\frac{1}{x}-1\right)\Psi(x)$, we will be able to determine $L$. Since $\left(\frac{1}{x}-1\right)$ is positive, the following inequalities are true $$-\dfrac{2\log(2)-1}{\log\left(x\right)}\left(\frac{1}{x}-1\right)\leq\left(\frac{1}{x}-1\right)\Psi(x)\leq-\dfrac{\log\left(1+x^{-1}\right)\left(1+x\right)-1}{{x\log\left(x\right)}}\left(\frac{1}{x}-1\right)$$ and now, since we know that $\lim\limits_{x\to1}\frac{1-1/x}{\log(x)}=1$, i.e. $\log(x)\sim 1-\frac{1}{x}$ around $1$, we finally see $$2\log(2)-1\leq \lim_{x\to1^-}\left(\frac{1}{x}-1\right)\Psi(x)\leq2\log(2)-1$$ by the squeeze theorem (that is, $\lim\limits_{x\to1^-}(1/x-1)\Psi(x)=2\log(2)-1$). With this in mind, we conclude that $$\log(L)=\lim_{x\to1^-}\log(L(x))=\lim_{x\to1^-}\left[-\frac{\log(2)}{x}+\left(\frac{1}{x}-1\right)\Psi(x)\right]=-\log(2)+2\log(2)-1=\log(2)-1$$ $$\boxed{\therefore L = \lim_{x\to1^-}\prod_{n=0}^\infty\left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}=e^{\log(2)-1}=\frac{2}{e}}$$ Hence, $\left\lceil\frac{1}{L}\right\rceil=\left\lceil\frac{e}{2}\right\rceil=\lceil1.3591409\dots\rceil=2$.