Evaluating $\lim_{x \to 1^{-}} \prod_{n=0}^{\infty} \left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}$ If $$\lim_{x \to 1^{-}} \prod_{n=0}^{\infty} \left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}=l$$ the question is to find out the integer part of the number $1/l$
I tried bringing the expression into telescopic series but failed to do so.Any ideas?Thanks.

Let's consider the logarithm of this product. The partial sum of the resulting series is
\begin{align}
\sum^n_{k=0}x^k\ln\frac{1+x^{k+1}}{1+x^k}&=\sum^n_{k=0}x^k\ln(1+x^{k+1})-\sum^n_{k=0}x^k\ln(1+x^k)
\\&=\sum^n_{k=0}x^k\ln(1+x^{k+1})-\sum^{n+1}_{k=1}x^k\ln(1+x^k)-\ln2+x^{n+1}\ln(1+x^{n+1})
\\&=\sum^n_{k=0}x^k\ln(1+x^{k+1})-\sum^{n}_{k=0}x^{k+1}\ln(1+x^{k+1})-\ln2+x^{n+1}\ln(1+x^{n+1})
\\&=\sum^n_{k=0}(x^k-x^{k+1})\ln(1+x^{k+1})-\ln2+x^{n+1}\ln(1+x^{n+1})
\end{align}
Now letting $n\to\infty$, we see
$$\sum^\infty_{k=0}x^k\ln\frac{1+x^{k+1}}{1+x^k}=\sum^\infty_{k=0}(x^k-x^{k+1})\ln(1+x^{k+1})-\ln2$$ for $x\in[0,1)$. But the series on the RHS is a Riemann sum for $\int^1_0\ln(1+u)\,du$ with a partition $u_k=x^k$, and $\max(u_k-u_{k+1})=1-x\to0$ as $x\to1-$, so the limit of the RHS is
$$\int^1_0\ln(1+u)\,du-\ln2=\ln2-1.$$ So the limit of the infinite product is
$$l=e^{\ln2-1}=\frac2e.$$
